Some Basic Concepts of chemistry

1. What is the molarity of a solution containing \(5.85\,\text{g}\) of \(\text{NaCl}\) dissolved in \(500\,\text{mL}\) of solution? [Molar mass of \(\text{NaCl} = 58.5\,\text{g/mol}\)]

Answer: 0.2 M

Explanation:
$$\text{Moles of NaCl} = \frac{5.85}{58.5} = 0.1\,\text{mol}$$ $$\text{Volume of solution} = \frac{500}{1000} = 0.5\,\text{L}$$ $$\text{Molarity} = \frac{0.1}{0.5} = 0.2\,\text{M}$$

2. Calculate the number of atoms present in \(4.6\,\text{g}\) of Sodium (\(\text{Na}\)). [Atomic mass: \(\text{Na}=23\,\text{u}\), \(N_A = 6.022 \times 10^{23}\)]

Answer: 1.2044 \times 10^{23}

Explanation:
$$\text{Moles of Na} = \frac{4.6}{23} = 0.2\,\text{mol}$$ $$\text{Number of atoms} = 0.2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{23}$$

3. How many significant figures are present in the number \(0.005600\)?

Answer: 4

Explanation:
Leading zeros before non-zero digits are non-significant, but trailing zeros after a decimal point are significant. The digits 5, 6, 0, and 0 are significant.

4. What mass of methane (\(\text{CH}_4\)) is required to produce \(22\,\text{g}\) of \(\text{CO}_2\) after complete combustion?

Answer: 8 g

Explanation:
Combustion equation: $$\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$$ \(1\,\text{mole of CH}_4 (16\,\text{g})\) produces \(1\,\text{mole of CO}_2 (44\,\text{g})\). $$\text{Mass of CH}_4 \text{ required} = \frac{16}{44} \times 22 = 8\,\text{g}$$

5. Determine the mass of magnesium required to produce \(220\,\text{mL}\) of hydrogen gas at STP upon reaction with excess dilute \(\text{HCl}\). [Molar mass of \(\text{Mg}=24\,\text{g/mol}\)]

Answer: 0.2357 g

Explanation:
Reaction: $$\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2$$ \(1\,\text{mole of Mg}\) produces \(22.4\,\text{L} (22400\,\text{mL})\) of \(\text{H}_2\). $$\text{Mass of Mg} = \frac{24}{22400} \times 220 \approx 0.2357\,\text{g}$$

6. Concentrated sulfuric acid is \(95\%\) \(\text{H}_2\text{SO}_4\) by weight with a density of \(1.834\,\text{g cm}^{-3}\). Calculate its molarity.

Answer: 17.8 M

Explanation:
$$1000\,\text{mL solution} = 1834\,\text{g}$$ $$\text{Mass of solute} = \frac{95}{100} \times 1834 = 1742.3\,\text{g}$$ $$\text{Molarity} = \frac{1742.3}{98} \approx 17.8\,\text{M}$$

7. What is the ratio of the number of molecules in a gaseous mixture containing oxygen and nitrogen in a mass ratio of \(1:4\)?

Answer: 7 : 32

Explanation:
$$\text{Ratio of moles} = \frac{w/32}{4w/28} = \frac{w}{32} \times \frac{28}{4w} = \frac{28}{128} = \frac{7}{32}$$

8. Find the mole fraction of solute in a \(1.00\,\text{m}\) aqueous solution.

Answer: 0.0177

Explanation:
\(1\,\text{m}\) solution means \(1\,\text{mole}\) of solute in \(1000\,\text{g}\) of water. $$\text{Moles of water} = \frac{1000}{18} = 55.55\,\text{mol}$$ $$\text{Mole fraction of solute} = \frac{1}{1 + 55.55} = 0.0177$$

9. How many moles of oxygen atoms are present in \(0.5\,\text{moles}\) of \(\text{CaCO}_3\)?

Answer: 1.5 moles

Explanation:
\(1\,\text{mole of CaCO}_3\) contains \(3\,\text{moles of oxygen atoms}\). $$\text{Moles of O} = 0.5 \times 3 = 1.5\,\text{moles}$$

10. What volume of \(0.1\,\text{M HCl}\) is required to neutralize \(25\,\text{mL}\) of \(0.2\,\text{M NaOH}\) solution?

Answer: 50 mL

Explanation:
$$M_1 V_1 = M_2 V_2$$ $$0.1 \times V_1 = 0.2 \times 25 \implies V_1 = 50\,\text{mL}$$

11. Calculate the mass percent of carbon in ethanol (\(\text{C}_2\text{H}_5\text{OH}\)). [Molar mass = \(46\,\text{g/mol}\)]

Answer: 52.17%

Explanation:
$$\text{Mass of carbon} = 2 \times 12 = 24\,\text{g}$$ $$\text{Mass percent} = \frac{24}{46} \times 100 = 52.17\%$$

12. If \(56.0\,\text{L}\) of nitrogen gas is mixed with excess hydrogen, calculate the volume of \(\text{NH}_3\) produced at STP assuming complete conversion.

Answer: 112 L

Explanation:
$$\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$$ \(1\,\text{vol of N}_2\) gives \(2\,\text{vol of NH}_3\). $$\text{Volume of NH}_3 = 2 \times 56.0 = 112\,\text{L}$$

13. Find the empirical formula of a compound containing \(40\%\) Carbon, \(6.67\%\) Hydrogen, and \(53.33\%\) Oxygen.

Answer: CH2O

Explanation:
Moles: \(\text{C} = \frac{40}{12} = 3.33\), \(\text{H} = \frac{6.67}{1} = 6.67\), \(\text{O} = \frac{53.33}{16} = 3.33\) Ratio: \(\text{C}:\text{H}:\text{O} = 1:2:1\). Empirical formula = \(\text{CH}_2\text{O}\).

14. What is the total number of electrons in \(1.6\,\text{g}\) of methane (\(\text{CH}_4\))?

Answer: 6.022 \times 10^{23}

Explanation:
Moles of \(\text{CH}_4 = \frac{1.6}{16} = 0.1\,\text{mol}\). \(1\,\text{molecule of CH}_4\) has \(6 + 4 = 10\) electrons. $$\text{Total electrons} = 0.1 \times 10 \times N_A = 1 \times N_A = 6.022 \times 10^{23}$$

15. What volume is occupied by \(3.011 \times 10^{23}\) molecules of \(\text{N}_2\) gas at STP?

Answer: 11.2 L

Explanation:
$$\text{Moles} = \frac{3.011 \times 10^{23}}{6.022 \times 10^{23}} = 0.5\,\text{mol}$$ $$\text{Volume at STP} = 0.5 \times 22.4 = 11.2\,\text{L}$$

16. Calculate the molarity of pure water at \(4^\circ\text{C}\) (density = \(1\,\text{g/mL}\)).

Answer: 55.56 M

Explanation:
Mass of \(1\,\text{L}\) water \(= 1000\,\text{g}\). $$\text{Moles of water} = \frac{1000}{18} = 55.56\,\text{mol}$$ $$\text{Molarity} = 55.56\,\text{M}$$

17. How many moles of \(\text{Al}_2\text{O}_3\) are present in \(51\,\text{g}\) of pure aluminum oxide? [Molar mass = \(102\,\text{g/mol}\)]

Answer: 0.5 moles

Explanation:
$$\text{Moles} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{51}{102} = 0.5\,\text{mol}$$

18. If \(0.5\,\text{moles}\) of \(\text{BaCl}_2\) reacts with \(0.2\,\text{moles}\) of \(\text{Na}_3\text{PO}_4\), maximum moles of \(\text{Ba}_3(\text{PO}_4)_2\) formed will be:

Answer: 0.1 moles

Explanation:
$$3\text{BaCl}_2 + 2\text{Na}_3\text{PO}_4 \rightarrow \text{Ba}_3(\text{PO}_4)_2 + 6\text{NaCl}$$ Here, \(\text{Na}_3\text{PO}_4\) is limiting. \(2\,\text{moles}\) form \(1\,\text{mole}\), so \(0.2\,\text{moles}\) form \(0.1\,\text{mole}\).

19. Convert \(27^\circ\text{C}\) into Kelvin scale.

Answer: 300.15 K

Explanation:
$$\text{Temperature in K} = 27 + 273.15 = 300.15\,\text{K}$$

20. Express \(0.0000456\) in scientific notation.

Answer: 4.56 \times 10^{-5}

Explanation:
Moving the decimal point 5 places to the right gives \(4.56 \times 10^{-5}\).

21. How many atoms are present in \(1\,\text{mole}\) of \(\text{NH}_3\)?

Answer: 2.4088 \times 10^{24}

Explanation:
\(1\,\text{molecule of NH}_3\) contains 4 atoms. $$\text{Total atoms} = 4 \times 6.022 \times 10^{23} = 2.4088 \times 10^{24}$$

22. Calculate the molality of a solution containing \(20\,\text{g}\) of \(\text{NaOH}\) in \(500\,\text{g}\) of water.

Answer: 1.0 m

Explanation:
$$\text{Moles of NaOH} = \frac{20}{40} = 0.5\,\text{mol}$$ $$\text{Molality} = \frac{0.5}{0.5\,\text{kg}} = 1.0\,\text{m}$$

23. What is the mass of \(1\,\text{atom of Carbon-12}\) in grams?

Answer: 1.99 \times 10^{-23} g

Explanation:
$$\text{Mass} = \frac{12}{6.022 \times 10^{23}} \approx 1.99 \times 10^{-23}\,\text{g}$$

24. A solution is prepared by mixing \(250\,\text{mL}\) of \(0.5\,\text{M HCl}\) and \(500\,\text{mL}\) of \(0.2\,\text{M HCl}\). Calculate resultant molarity.

Answer: 0.3 M

Explanation:
$$M_{\text{mix}} = \frac{M_1 V_1 + M_2 V_2}{V_1 + V_2} = \frac{(0.5 \times 250) + (0.2 \times 500)}{250 + 500} = \frac{125 + 100}{750} = 0.3\,\text{M}$$

25. What is the law illustrated by the formation of \(\text{CO}\) and \(\text{CO}_2\) from carbon and oxygen?

Answer: Law of Multiple Proportions

Explanation:
When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers.

26. How many grams of \(\text{CaO}\) are obtained by heating \(20\,\text{g}\) of pure \(\text{CaCO}_3\)?

Answer: 11.2 g

Explanation:
$$\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$$ \(100\,\text{g CaCO}_3\) gives \(56\,\text{g CaO}\). $$\text{Mass of CaO} = \frac{56}{100} \times 20 = 11.2\,\text{g}$$

27. Round off \(10.3456\) to 3 significant figures.

Answer: 10.3

Explanation:
The fourth digit is 4 (< 5), so rounding to 3 significant figures yields 10.3.

28. Find the number of moles of solute present in \(250\,\text{mL}\) of \(0.1\,\text{M KNO}_3\) solution.

Answer: 0.025 moles

Explanation:
$$\text{Moles} = \text{Molarity} \times \text{Volume in L} = 0.1 \times 0.25 = 0.025\,\text{mol}$$

29. What is the mass percent of oxygen in water (\(\text{H}_2\text{O}\))?

Answer: 88.89%

Explanation:
$$\%\text{ Oxygen} = \frac{16}{18} \times 100 = 88.89\%$$

30. Which contains maximum number of molecules? (A) \(1\,\text{g H}_2\) (B) \(1\,\text{g N}_2\) (C) \(1\,\text{g O}_2\) (D) \(1\,\text{g CH}_4\)

Answer: 1 g H2

Explanation:
Moles \(= \text{Mass} / \text{Molar mass}\). \(\text{H}_2\) has the smallest molar mass (\(2\,\text{g/mol}\)), yielding the highest number of moles (\(0.5\,\text{mol}\)).

31. Calculate the density of \(\text{CO}_2\) gas at STP in \(\text{g/L}\).

Answer: 1.96 g/L

Explanation:
$$\text{Density} = \frac{\text{Molar mass}}{\text{Molar volume}} = \frac{44}{22.4} \approx 1.96\,\text{g/L}$$

32. How many moles of \(\text{HCl}\) are required to react completely with \(1\,\text{mole}\) of \(\text{Mg(OH)}_2\)?

Answer: 2 moles

Explanation:
$$\text{Mg(OH)}_2 + 2\text{HCl} \rightarrow \text{MgCl}_2 + 2\text{H}_2\text{O}$$

33. State the total number of neutrons in \(7\,\text{g}\) of \(^{14}\text{N}^{3-}\) ions.

Answer: 2.107 \times 10^{24}

Explanation:
Moles \(= \frac{7}{14} = 0.5\,\text{mol}\). Each nitrogen nucleus has 7 neutrons. $$\text{Total neutrons} = 0.5 \times 7 \times N_A = 3.5 N_A = 2.107 \times 10^{24}$$

34. What is the normality of \(0.5\,\text{M H}_2\text{SO}_4\) solution?

Answer: 1.0 N

Explanation:
$$\text{Normality} = \text{Molarity} \times n\text{-factor} = 0.5 \times 2 = 1.0\,\text{N}$$

35. Determine the volume of \(0.2\,\text{M NaOH}\) required to neutralize \(40\,\text{mL}\) of \(0.1\,\text{M H}_2\text{SO}_4\).

Answer: 40 mL

Explanation:
$$N_1 V_1 = N_2 V_2 \implies (0.1 \times 2) \times 40 = 0.2 \times V_2 \implies V_2 = 40\,\text{mL}$$

36. If molecular formula of a compound is \(\text{C}_6\text{H}_{12}\text{O}_6\), what is its empirical formula?

Answer: CH2O

Explanation:
Simplest whole number ratio of \(\text{C:H:O}\) is \(1:2:1\).

37. Calculate the mass of \(0.1\,\text{mole}\) of glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)).

Answer: 18 g

Explanation:
$$\text{Mass} = 0.1 \times 180 = 18\,\text{g}$$

38. What is the SI unit of luminous intensity?

Answer: Candela (cd)

Explanation:
Candela is the standard SI base unit for luminous intensity.

39. The mole fraction of glucose in an aqueous solution is \(0.1\). Find the molality of the solution.

Answer: 6.17 m

Explanation:
$$\text{Molality} = \frac{\chi_{\text{solute}} \times 1000}{\chi_{\text{solvent}} \times M_{\text{solvent}}} = \frac{0.1 \times 1000}{0.9 \times 18} = 6.17\,\text{m}$$

40. How many total atoms are present in \(18\,\text{mg}\) of water?

Answer: 1.806 \times 10^{21}

Explanation:
$$\text{Moles} = \frac{0.018}{18} = 10^{-3}\,\text{mol}$$ $$\text{Atoms} = 3 \times 10^{-3} \times 6.022 \times 10^{23} = 1.806 \times 10^{21}$$

41. How many moles of \(\text{O}_2\) are required for the complete combustion of \(1\,\text{mole}\) of propane (\(\text{C}_3\text{H}_8\))?

Answer: 5 moles

Explanation:
$$\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$$

42. Express \(1\,\text{amu}\) in grams.

Answer: 1.66 \times 10^{-24} g

Explanation:
$$1\,\text{amu} = \frac{1}{N_A}\,\text{g} = \frac{1}{6.022 \times 10^{23}} = 1.66 \times 10^{-24}\,\text{g}$$

43. Calculate the number of moles of \(\text{Na}^+\) ions in \(200\,\text{mL}\) of \(0.5\,\text{M Na}_2\text{SO}_4\) solution.

Answer: 0.2 moles

Explanation:
$$\text{Moles of Na}_2\text{SO}_4 = 0.5 \times 0.2 = 0.1\,\text{mol}$$ $$\text{Moles of Na}^+ = 2 \times 0.1 = 0.2\,\text{mol}$$

44. What is the mass of \(2.24\,\text{L}\) of \(\text{CO}_2\) gas at STP?

Answer: 4.4 g

Explanation:
$$\text{Moles} = \frac{2.24}{22.4} = 0.1\,\text{mol}$$ $$\text{Mass} = 0.1 \times 44 = 4.4\,\text{g}$$

45. What is the equivalent mass of \(\text{H}_3\text{PO}_4\) in a complete neutralization reaction? [Molar mass = \(M\)]

Answer: M / 3

Explanation:
For complete neutralization, \(\text{H}_3\text{PO}_4\) releases 3 protons (\(n\text{-factor} = 3\)). $$\text{Equivalent Mass} = \frac{M}{3}$$

46. Calculate the percentage of water of crystallization in washing soda (\(\text{Na}_2\text{CO}_3\cdot 10\text{H}_2\text{O}\)). [Molar mass = \(286\,\text{g/mol}\)]

Answer: 62.94%

Explanation:
$$\text{Mass of 10H}_2\text{O} = 10 \times 18 = 180\,\text{g}$$ $$\%\text{ Water} = \frac{180}{286} \times 100 = 62.94\%$$

47. What is the total mass of reactants and products conserved according to which law?

Answer: Law of Conservation of Mass

Explanation:
The Law of Conservation of Mass states that matter can neither be created nor destroyed in a chemical reaction.

48. Find the volume of oxygen gas required at STP to completely burn \(10\,\text{L}\) of \(\text{CO}\) gas.

Answer: 5 L

Explanation:
$$2\text{CO}(g) + \text{O}_2(g) \rightarrow 2\text{CO}_2(g)$$ \(2\,\text{vol of CO}\) requires \(1\,\text{vol of O}_2\). Therefore, \(10\,\text{L}\) requires \(5\,\text{L}\).

49. Calculate the number of molecules present in one drop of water weighing \(0.05\,\text{g}\).

Answer: 1.67 \times 10^{21}

Explanation:
$$\text{Moles} = \frac{0.05}{18} = 0.00277\,\text{mol}$$ $$\text{Molecules} = 0.00277 \times 6.022 \times 10^{23} = 1.67 \times 10^{21}$$

50. What is the empirical formula mass of glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\))?

Answer: 30 u

Explanation:
Empirical formula of glucose is \(\text{CH}_2\text{O}\). $$\text{Empirical mass} = 12 + (2 \times 1) + 16 = 30\,\text{u}$$

51. How many moles of electrons are present in \(1.6\,\text{g}\) of methane (\(\text{CH}_4\))?

Answer: 1.0 mole

Explanation:
$$\text{Moles of CH}_4 = \frac{1.6}{16} = 0.1\,\text{mol}$$ Each molecule of \(\text{CH}_4\) contains \(6 + 4 = 10\) electrons. $$\text{Moles of electrons} = 0.1 \times 10 = 1.0\,\text{mole}$$

52. What volume of \(0.5\,\text{M H}_2\text{SO}_4\) is required to neutralize \(100\,\text{mL}\) of \(1\,\text{M NaOH}\)?

Answer: 100 mL

Explanation:
$$N_1 V_1 = N_2 V_2$$ $$\text{Normality of H}_2\text{SO}_4 = 0.5 \times 2 = 1\,\text{N}$$ $$1 \times V_1 = 1 \times 100 \implies V_1 = 100\,\text{mL}$$

53. Calculate the number of chloride ions present in \(100\,\text{mL}\) of \(0.1\,\text{M CaCl}_2\) solution.

Answer: 1.2044 \times 10^{22}

Explanation:
$$\text{Moles of CaCl}_2 = 0.1 \times 0.1 = 0.01\,\text{mol}$$ $$\text{Moles of Cl}^- = 2 \times 0.01 = 0.02\,\text{mol}$$ $$\text{Number of Cl}^- \text{ ions} = 0.02 \times 6.022 \times 10^{23} = 1.2044 \times 10^{22}$$

54. What is the mass of oxygen present in \(0.25\,\text{moles}\) of \(\text{MgSO}_4\cdot 7\text{H}_2\text{O}\)?

Answer: 44 g

Explanation:
\(1\,\text{mole of MgSO}_4\cdot 7\text{H}_2\text{O}\) contains \(4 + 7 = 11\,\text{moles of O atoms}\). $$\text{Moles of O} = 0.25 \times 11 = 2.75\,\text{mol}$$ $$\text{Mass of O} = 2.75 \times 16 = 44\,\text{g}$$

55. Calculate the molality of an aqueous solution of urea in which the mole fraction of urea is \(0.02\).

Answer: 1.13 m

Explanation:
$$\text{Molality} = \frac{\chi_{\text{solute}} \times 1000}{\chi_{\text{solvent}} \times M_{\text{solvent}}} = \frac{0.02 \times 1000}{0.98 \times 18} \approx 1.13\,\text{m}$$

56. Which law states that equal volumes of all gases under the same conditions of temperature and pressure contain equal numbers of molecules?

Answer: Avogadro's Law

Explanation:
Avogadro's law specifies that \(V \propto n\) at constant temperature and pressure.

57. How many significant figures are there in \(1.0070\)?

Answer: 5

Explanation:
Zeros between non-zero digits and trailing zeros after a decimal point are significant. Thus, all 5 digits are significant.

58. What mass of \(\text{CO}_2\) is produced when \(10\,\text{g}\) of pure \(\text{CaCO}_3\) is treated with excess \(\text{HCl}\)?

Answer: 4.4 g

Explanation:
$$\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2$$ \(100\,\text{g CaCO}_3\) yields \(44\,\text{g CO}_2\). $$\text{Mass of CO}_2 = \frac{44}{100} \times 10 = 4.4\,\text{g}$$

59. Find the mass of one molecule of \(\text{water}\) (\(\text{H}_2\text{O}\)) in grams.

Answer: 2.99 \times 10^{-23} g

Explanation:
$$\text{Mass} = \frac{18}{6.022 \times 10^{23}} \approx 2.99 \times 10^{-23}\,\text{g}$$

60. What is the mass percentage of nitrogen in urea (\(\text{NH}_2\text{CONH}_2\))? [Molar mass = \(60\,\text{g/mol}\)]

Answer: 46.67%

Explanation:
$$\text{Mass of N} = 2 \times 14 = 28\,\text{g}$$ $$\%\text{ Nitrogen} = \frac{28}{60} \times 100 = 46.67\%$$

61. Calculate the volume of \(0.5\,\text{M NaOH}\) required to react completely with \(50\,\text{mL}\) of \(0.2\,\text{M H}_3\text{PO}_4\).

Answer: 60 mL

Explanation:
$$\text{Milli-equivalents of acid} = 50 \times 0.2 \times 3 = 30$$ $$\text{Milli-equivalents of base} = V \times 0.5 \times 1 = 0.5 V$$ $$0.5 V = 30 \implies V = 60\,\text{mL}$$

62. Determine the empirical formula of an oxide of iron which has \(69.9\%\) iron and \(30.1\%\) oxygen by mass. [Atomic mass: \(\text{Fe}=55.85\), \(\text{O}=16.0\)]

Answer: Fe2O3

Explanation:
Moles of \(\text{Fe} = \frac{69.9}{55.85} = 1.25\); Moles of \(\text{O} = \frac{30.1}{16} = 1.88\).
Ratio \(= \frac{1.25}{1.25} : \frac{1.88}{1.25} = 1 : 1.5 = 2 : 3\). Empirical formula is \(\text{Fe}_2\text{O}_3\).

63. What volume of oxygen at STP is liberated when \(0.1\,\text{mole}\) of \(\text{KClO}_3\) is heated completely?

Answer: 3.36 L

Explanation:
$$2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2$$ \(2\,\text{moles KClO}_3\) yield \(3\,\text{moles O}_2\). $$\text{Moles of O}_2 = 0.1 \times \frac{3}{2} = 0.15\,\text{mol}$$ $$\text{Volume at STP} = 0.15 \times 22.4 = 3.36\,\text{L}$$

64. Calculate the total number of protons present in \(10\,\text{g}\) of \(\text{CaCO}_3\).

Answer: 3.011 \times 10^{24}

Explanation:
Moles \(= \frac{10}{100} = 0.1\,\text{mol}\). Protons in 1 unit of \(\text{CaCO}_3 = 20 (\text{Ca}) + 6 (\text{C}) + 3\times8 (\text{O}) = 50\). $$\text{Total protons} = 0.1 \times 50 \times 6.022 \times 10^{23} = 3.011 \times 10^{24}$$

65. If the density of a \(3\,\text{M}\) solution of \(\text{NaCl}\) is \(1.25\,\text{g mL}^{-1}\), calculate its molality. [Molar mass of \(\text{NaCl} = 58.5\,\text{g/mol}\)]

Answer: 2.79 m

Explanation:
Mass of \(1\,\text{L solution} = 1250\,\text{g}\). Mass of \(\text{NaCl} = 3 \times 58.5 = 175.5\,\text{g}\). Mass of solvent \(= 1250 - 175.5 = 1074.5\,\text{g} = 1.0745\,\text{kg}\). $$\text{Molality} = \frac{3}{1.0745} \approx 2.79\,\text{m}$$

66. How many moles of \(\text{KMnO}_4\) are reduced by \(1\,\text{mole}\) of \(\text{FeSO}_4\) in acidic medium?

Answer: 0.2 moles

Explanation:
$$\text{Equivalents of KMnO}_4 = \text{Equivalents of FeSO}_4$$ \(n\)-factor of \(\text{KMnO}_4\) in acid \(= 5\); \(n\)-factor of \(\text{FeSO}_4 = 1\). $$n_{\text{KMnO}_4} \times 5 = 1 \times 1 \implies n_{\text{KMnO}_4} = 0.2\,\text{mol}$$

67. Express \(0.00350\) in standard exponential form.

Answer: 3.50 \times 10^{-3}

Explanation:
Shift decimal 3 places to the right: \(3.50 \times 10^{-3}\).

68. What is the number of moles of hydrogen atoms in \(0.2\,\text{moles}\) of \(\text{C}_6\text{H}_{12}\text{O}_6\)?

Answer: 2.4 moles

Explanation:
\(1\,\text{mole of glucose}\) has \(12\,\text{moles of H}\). $$\text{Moles of H} = 0.2 \times 12 = 2.4\,\text{moles}$$

69. How many grams of \(\text{NaOH}\) are present in \(250\,\text{mL}\) of a \(0.5\,\text{M}\) solution?

Answer: 5.0 g

Explanation:
$$\text{Moles} = 0.5 \times 0.25 = 0.125\,\text{mol}$$ $$\text{Mass} = 0.125 \times 40 = 5.0\,\text{g}$$

70. What is the mole fraction of \(\text{C}_2\text{H}_5\text{OH}\) in a solution made by mixing \(46\,\text{g}\) ethanol and \(54\,\text{g}\) water?

Answer: 0.25

Explanation:
Moles of ethanol \(= \frac{46}{46} = 1\,\text{mol}\). Moles of water \(= \frac{54}{18} = 3\,\text{mol}\). $$\text{Mole fraction of ethanol} = \frac{1}{1 + 3} = 0.25$$

71. How many molecules of \(\text{CO}_2\) are formed when \(0.5\,\text{mole}\) of carbon burns completely in air?

Answer: 3.011 \times 10^{23}

Explanation:
$$\text{C} + \text{O}_2 \rightarrow \text{CO}_2$$ \(0.5\,\text{mole C}\) produces \(0.5\,\text{mole CO}_2\). $$\text{Molecules} = 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}$$

72. State the law of definite proportions (or constant composition).

Answer: Joseph Proust

Explanation:
The Law of Definite Proportions was formulated by Joseph Proust in 1799.

73. What is the density of a gas at STP if its vapor density relative to hydrogen is \(14\)?

Answer: 1.25 g/L

Explanation:
$$\text{Molar mass} = 2 \times \text{Vapor Density} = 2 \times 14 = 28\,\text{g/mol}$$ $$\text{Density at STP} = \frac{28}{22.4} = 1.25\,\text{g/L}$$

74. Calculate the percentage of sulfur in barium sulfate (\(\text{BaSO}_4\)). [Atomic mass: \(\text{Ba}=137\), \(\text{S}=32\), \(\text{O}=16\)]

Answer: 13.73%

Explanation:
$$\text{Molar mass of BaSO}_4 = 137 + 32 + 64 = 233\,\text{g/mol}$$ $$\%\text{ Sulfur} = \frac{32}{233} \times 100 = 13.73\%$$

75. What volume of water must be added to \(200\,\text{mL}\) of \(0.5\,\text{M HCl}\) to make it a \(0.1\,\text{M}\) solution?

Answer: 800 mL

Explanation:
$$M_1 V_1 = M_2 V_2 \implies 0.5 \times 200 = 0.1 \times V_2 \implies V_2 = 1000\,\text{mL}$$ $$\text{Water added} = 1000 - 200 = 800\,\text{mL}$$

76. Calculate the molar mass of a gas if \(0.28\,\text{g}\) of it occupies \(224\,\text{mL}\) at STP.

Answer: 28 g/mol

Explanation:
$$\text{Moles} = \frac{224}{22400} = 0.01\,\text{mol}$$ $$\text{Molar mass} = \frac{0.28}{0.01} = 28\,\text{g/mol}$$

77. How many subatomic particles (protons, neutrons, and electrons) are in \(1\,\text{mole}\) of \(^1_1\text{H}\) atoms?

Answer: 1.2044 \times 10^{24}

Explanation:
A \(^1_1\text{H}\) atom has 1 proton and 1 electron (0 neutrons) = 2 particles. $$\text{Total particles} = 2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{24}$$

78. Find the empirical formula of a hydrocarbon containing \(85.7\%\) Carbon and \(14.3\%\) Hydrogen.

Answer: CH2

Explanation:
Moles of \(\text{C} = \frac{85.7}{12} = 7.14\); Moles of \(\text{H} = \frac{14.3}{1} = 14.3\).
Ratio \(\text{C:H} = 1:2\). Empirical formula = \(\text{CH}_2\).

79. What is the equivalent mass of \(\text{KMnO}_4\) in a strongly alkaline medium? [Molar mass = \(M\)]

Answer: M

Explanation:
In a strongly alkaline medium, \(\text{MnO}_4^- + e^- \rightarrow \text{MnO}_4^{2-}\) (\(n\text{-factor} = 1\)). $$\text{Equivalent Mass} = \frac{M}{1} = M$$

80. How many moles of \(\text{Na}^+\) are in a solution containing \(5.85\,\text{g}\) \(\text{NaCl}\) and \(14.2\,\text{g}\) \(\text{Na}_2\text{SO}_4\)?

Answer: 0.3 moles

Explanation:
Moles of \(\text{Na}^+\) from \(\text{NaCl} = \frac{5.85}{58.5} = 0.1\,\text{mol}\).
Moles of \(\text{Na}^+\) from \(\text{Na}_2\text{SO}_4 = 2 \times \frac{14.2}{142} = 0.2\,\text{mol}\).
$$\text{Total Na}^+ = 0.1 + 0.2 = 0.3\,\text{mol}$$

81. What mass of \(\text{H}_2\text{O}\) contains the same number of molecules as \(44\,\text{g}\) of \(\text{CO}_2\)?

Answer: 18 g

Explanation:
\(44\,\text{g CO}_2 = 1\,\text{mole CO}_2\). Equal molecules means equal moles, so \(1\,\text{mole H}_2\text{O} = 18\,\text{g}\).

82. The result of \(2.34 \times 1.2\) rounded to correct significant figures is:

Answer: 2.8

Explanation:
\(2.34 \times 1.2 = 2.808\). The factor with the fewest significant figures is \(1.2\) (2 sig figs). Rounding to 2 significant figures gives \(2.8\).

83. How many moles of \(\text{O}_2\) are required to react completely with \(27\,\text{g}\) of Aluminum? [Atomic mass of \(\text{Al} = 27\)]

Answer: 0.75 moles

Explanation:
$$4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3$$ \(1\,\text{mole of Al} (27\,\text{g})\) requires \(\frac{3}{4} = 0.75\,\text{moles of O}_2\).

84. What is the normality of a solution obtained by mixing \(100\,\text{mL}\) of \(0.2\,\text{N HCl}\) and \(300\,\text{mL}\) of \(0.1\,\text{N H}_2\text{SO}_4\)?

Answer: 0.125 N

Explanation:
$$N_{\text{mix}} = \frac{N_1 V_1 + N_2 V_2}{V_1 + V_2} = \frac{(0.2 \times 100) + (0.1 \times 300)}{100 + 300} = \frac{20 + 30}{400} = 0.125\,\text{N}$$

85. Calculate the mass of anhydrous \(\text{Na}_2\text{CO}_3\) present in \(250\,\text{mL}\) of a \(0.1\,\text{M}\) solution.

Answer: 2.65 g

Explanation:
$$\text{Moles} = 0.1 \times 0.25 = 0.025\,\text{mol}$$ $$\text{Mass} = 0.025 \times 106 = 2.65\,\text{g}$$

86. Find the number of moles of gas present in a volume of \(5.6\,\text{dm}^3\) at STP.

Answer: 0.25 moles

Explanation:
\(1\,\text{dm}^3 = 1\,\text{L}\). $$\text{Moles} = \frac{5.6}{22.4} = 0.25\,\text{mol}$$

87. What is the vapor density of \(\text{SO}_2\) gas relative to hydrogen? [Atomic mass: \(\text{S}=32\), \(\text{O}=16\)]

Answer: 32

Explanation:
$$\text{Molar mass of SO}_2 = 32 + 32 = 64\,\text{g/mol}$$ $$\text{Vapor Density} = \frac{\text{Molar mass}}{2} = \frac{64}{2} = 32$$

88. What is the mass of \(6.022 \times 10^{22}\) molecules of nitrogen gas (\(\text{N}_2\))?

Answer: 2.8 g

Explanation:
$$\text{Moles} = \frac{6.022 \times 10^{22}}{6.022 \times 10^{23}} = 0.1\,\text{mol}$$ $$\text{Mass} = 0.1 \times 28 = 2.8\,\text{g}$$

89. State the unit of molality.

Answer: mol/kg

Explanation:
Molality is defined as moles of solute per kilogram of solvent (\(\text{mol kg}^{-1}\)).

90. How many moles of \(\text{CO}_2\) are formed when \(1\,\text{mole}\) of \(\text{CH}_4\) reacts with \(1\,\text{mole}\) of \(\text{O}_2\)?

Answer: 0.5 moles

Explanation:
$$\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$$ \(\text{O}_2\) is the limiting reagent. \(2\,\text{moles O}_2\) form \(1\,\text{mole CO}_2\), so \(1\,\text{mole O}_2\) forms \(0.5\,\text{moles CO}_2\).

91. What is the mass percent of hydrogen in methane (\(\text{CH}_4\))?

Answer: 25%

Explanation:
$$\%\text{ Hydrogen} = \frac{4}{16} \times 100 = 25\%$$

92. Calculate the volume of \(0.1\,\text{M Ba(OH)}_2\) needed to neutralize \(20\,\text{mL}\) of \(0.1\,\text{M HCl}\).

Answer: 10 mL

Explanation:
$$\text{Milli-equivalents of acid} = 20 \times 0.1 \times 1 = 2$$ $$\text{Milli-equivalents of base} = V \times 0.1 \times 2 = 0.2 V$$ $$0.2 V = 2 \implies V = 10\,\text{mL}$$

93. Round off \(0.04597\) to 3 significant figures.

Answer: 0.0460

Explanation:
The 4th significant digit is 7, so 9 rounds up: \(0.0460\).

94. How many oxygen atoms are in \(4.4\,\text{g}\) of \(\text{CO}_2\)?

Answer: 1.2044 \times 10^{23}

Explanation:
$$\text{Moles of CO}_2 = \frac{4.4}{44} = 0.1\,\text{mol}$$ $$\text{Moles of O atoms} = 0.1 \times 2 = 0.2\,\text{mol}$$ $$\text{Number of O atoms} = 0.2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{23}$$

95. What is the concentration of a solution in \(\text{g/L}\) if its molarity is \(0.2\,\text{M}\) and molar mass of solute is \(50\,\text{g/mol}\)?

Answer: 10 g/L

Explanation:
$$\text{Strength (g/L)} = \text{Molarity} \times \text{Molar mass} = 0.2 \times 50 = 10\,\text{g/L}$$

96. What is the ratio of empirical formula mass to molecular formula mass for benzene (\(\text{C}_6\text{H}_6\))?

Answer: 1 : 6

Explanation:
Empirical formula is \(\text{CH}\) (mass = \(13\)). Molecular formula is \(\text{C}_6\text{H}_6\) (mass = \(78\)). $$\text{Ratio} = \frac{13}{78} = \frac{1}{6}$$

97. Calculate the mass of sodium produced when \(1\,\text{mole}\) of \(\text{NaCl}\) is completely electrolyzed.

Answer: 23 g

Explanation:
\(1\,\text{mole of NaCl}\) yields \(1\,\text{mole of Na}\) atoms (\(23\,\text{g}\)).

98. What volume is occupied by \(16\,\text{g}\) of oxygen gas (\(\text{O}_2\)) at STP?

Answer: 11.2 L

Explanation:
$$\text{Moles} = \frac{16}{32} = 0.5\,\text{mol}$$ $$\text{Volume} = 0.5 \times 22.4 = 11.2\,\text{L}$$

99. How many moles of water are produced by the complete neutralization of \(1\,\text{mole}\) of \(\text{H}_2\text{SO}_4\) with \(\text{NaOH}\)?

Answer: 2 moles

Explanation:
$$\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$$

100. Calculate the number of moles of solute in \(500\,\text{mL}\) of a \(2\,\text{M}\) solution.

Answer: 1.0 mole

Explanation:
$$\text{Moles} = \text{Molarity} \times \text{Volume in L} = 2 \times 0.5 = 1.0\,\text{mole}$$

101. Calculate the number of moles of hydrogen gas evolved when \(5.4\,\text{g}\) of aluminum reacts completely with excess aqueous \(\text{NaOH}\). [Atomic mass of \(\text{Al} = 27\)]

Answer: 0.3 moles

Explanation:
Reaction: $$2\text{Al} + 2\text{NaOH} + 6\text{H}_2\text{O} \rightarrow 2\text{Na}[\text{Al(OH)}_4] + 3\text{H}_2$$ Moles of \(\text{Al} = \frac{5.4}{27} = 0.2\,\text{mol}\).
\(2\,\text{moles of Al}\) yield \(3\,\text{moles of H}_2\). $$\text{Moles of H}_2 = 0.2 \times \frac{3}{2} = 0.3\,\text{moles}$$

102. What is the concentration of \(\text{Cl}^-\) ions in \(\text{mol/L}\) when \(200\,\text{mL}\) of \(0.1\,\text{M NaCl}\) is mixed with \(300\,\text{mL}\) of \(0.2\,\text{M MgCl}_2\)?

Answer: 0.28 M

Explanation:
Moles of \(\text{Cl}^-\) from \(\text{NaCl} = 0.2 \times 0.1 = 0.02\,\text{mol}\).
Moles of \(\text{Cl}^-\) from \(\text{MgCl}_2 = 0.3 \times 0.2 \times 2 = 0.12\,\text{mol}\).
Total moles of \(\text{Cl}^- = 0.02 + 0.12 = 0.14\,\text{mol}\).
$$\text{Concentration} = \frac{0.14\,\text{mol}}{0.5\,\text{L}} = 0.28\,\text{M}$$

103. How many significant figures are there in the result of the calculation \(12.11 + 18.0 + 1.012\)?

Answer: 3

Explanation:
\(12.11 + 18.0 + 1.012 = 31.122\).
In addition, the result must be rounded to the least number of decimal places in any of the numbers (\(18.0\) has 1 decimal place). Thus, the result is rounded to \(31.1\), which contains 3 significant figures.

104. Calculate the percentage of water of crystallization in copper sulfate pentahydrate (\(\text{CuSO}_4\cdot 5\text{H}_2\text{O}\)). [Molar mass = \(249.5\,\text{g/mol}\)]

Answer: 36.07%

Explanation:
$$\text{Mass of 5H}_2\text{O} = 5 \times 18 = 90\,\text{g}$$ $$\%\text{ Water} = \frac{90}{249.5} \times 100 = 36.07\%$$

105. Find the total number of valence electrons present in \(4.2\,\text{g}\) of azide ion (\(\text{N}_3^-\)).

Answer: 1.6 \times N_A

Explanation:
Molar mass of \(\text{N}_3^- = 3 \times 14 = 42\,\text{g/mol}\).
Moles of \(\text{N}_3^- = \frac{4.2}{42} = 0.1\,\text{mol}\).
Valence electrons in 1 \(\text{N}_3^-\) ion \(= 3(5) + 1 = 16\).
$$\text{Total valence electrons} = 0.1 \times 16 \times N_A = 1.6 N_A$$

106. What mass of \(\text{HNO}_3\) is contained in \(50\,\text{mL}\) of a \(2\,\text{M}\) aqueous solution? [Molar mass = \(63\,\text{g/mol}\)]

Answer: 6.3 g

Explanation:
$$\text{Moles} = 2 \times 0.05 = 0.1\,\text{mol}$$ $$\text{Mass} = 0.1 \times 63 = 6.3\,\text{g}$$

107. A compound contains \(50\%\) element X (atomic mass \(= 10\)) and \(50\%\) element Y (atomic mass \(= 20\)). What is its empirical formula?

Answer: X2Y

Explanation:
Moles of X \(= \frac{50}{10} = 5\); Moles of Y \(= \frac{50}{20} = 2.5\).
Ratio X : Y \(= 5 : 2.5 = 2 : 1\). Empirical formula is \(\text{X}_2\text{Y}\).

108. Determine the volume of air (containing \(21\%\) \(\text{O}_2\) by volume) needed for complete combustion of \(10\,\text{L}\) of methane (\(\text{CH}_4\)) at STP.

Answer: 95.24 L

Explanation:
$$\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$$ \(10\,\text{L CH}_4\) requires \(2 \times 10 = 20\,\text{L O}_2\).
$$\text{Volume of air} = \frac{20}{0.21} \approx 95.24\,\text{L}$$

109. What is the ratio of number of atoms present in \(1\,\text{g}\) of \(\text{He}\) to \(1\,\text{g}\) of \(\text{O}_2\)? [Atomic mass: \(\text{He}=4\), \(\text{O}=16\)]

Answer: 4 : 1

Explanation:
Atoms of \(\text{He} = \frac{1}{4} \times N_A\).
Atoms of \(\text{O}_2 = \left(\frac{1}{32} \times N_A\right) \times 2 = \frac{1}{16} \times N_A\).
$$\text{Ratio} = \frac{1/4}{1/16} = \frac{16}{4} = 4 : 1$$

110. Find the equivalent mass of \(\text{K}_2\text{Cr}_2\text{O}_7\) in acidic medium. [Molar mass = \(M\)]

Answer: M / 6

Explanation:
$$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$$ The change in oxidation state per mole of \(\text{K}_2\text{Cr}_2\text{O}_7\) is 6 (\(n\text{-factor} = 6\)).
$$\text{Equivalent Mass} = \frac{M}{6}$$

111. How many moles of \(\text{P}_4\text{O}_{10}\) can be produced from the reaction of \(62\,\text{g}\) of phosphorus (\(\text{P}_4\)) with excess oxygen? [Molar mass of \(\text{P}_4 = 124\,\text{g/mol}\)]

Answer: 0.5 moles

Explanation:
$$\text{P}_4 + 5\text{O}_2 \rightarrow \text{P}_4\text{O}_{10}$$ Moles of \(\text{P}_4 = \frac{62}{124} = 0.5\,\text{mol}\).
\(1\,\text{mole of P}_4\) gives \(1\,\text{mole of P}_4\text{O}_{10}\), so \(0.5\,\text{mole}\) gives \(0.5\,\text{mole}\).

112. Calculate the molality of a \(10\%\) (w/w) aqueous solution of \(\text{NaOH}\).

Answer: 2.78 m

Explanation:
\(10\,\text{g NaOH}\) in \(90\,\text{g}\) water.
$$\text{Moles of NaOH} = \frac{10}{40} = 0.25\,\text{mol}$$ $$\text{Molality} = \frac{0.25}{0.09\,\text{kg}} \approx 2.78\,\text{m}$$

113. What mass of \(\text{AgCl}\) will be precipitated when \(100\,\text{mL}\) of \(0.1\,\text{M AgNO}_3\) is mixed with \(100\,\text{mL}\) of \(0.1\,\text{M NaCl}\)? [Molar mass of \(\text{AgCl} = 143.5\,\text{g/mol}\)]

Answer: 1.435 g

Explanation:
Moles of \(\text{AgNO}_3 = 0.1 \times 0.1 = 0.01\,\text{mol}\).
Moles of \(\text{NaCl} = 0.1 \times 0.1 = 0.01\,\text{mol}\).
Moles of \(\text{AgCl}\) formed \(= 0.01\,\text{mol}\).
$$\text{Mass of AgCl} = 0.01 \times 143.5 = 1.435\,\text{g}$$

114. Express the concentration of pure liquid ethanol (\(\text{C}_2\text{H}_5\text{OH}\), density \(= 0.789\,\text{g/mL}\)) in molarity. [Molar mass = \(46\,\text{g/mol}\)]

Answer: 17.15 M

Explanation:
Mass of \(1\,\text{L ethanol} = 789\,\text{g}\).
$$\text{Molarity} = \frac{\text{Moles}}{\text{Volume in L}} = \frac{789 / 46}{1} \approx 17.15\,\text{M}$$

115. How many oxygen atoms are present in \(50\,\text{g}\) of calcium carbonate (\(\text{CaCO}_3\))?

Answer: 9.033 \times 10^{23}

Explanation:
Moles of \(\text{CaCO}_3 = \frac{50}{100} = 0.5\,\text{mol}\).
Moles of \(\text{O atoms} = 0.5 \times 3 = 1.5\,\text{mol}\).
$$\text{Number of O atoms} = 1.5 \times 6.022 \times 10^{23} = 9.033 \times 10^{23}$$

116. If \(100\,\text{mL}\) of a gas at STP weighs \(0.179\,\text{g}\), what is the molecular mass of the gas?

Answer: 40.1 g/mol

Explanation:
$$\text{Moles} = \frac{100}{22400} = 0.004464\,\text{mol}$$ $$\text{Molar mass} = \frac{0.179}{0.004464} \approx 40.1\,\text{g/mol}$$

117. What volume of \(12\,\text{M HCl}\) is required to prepare \(2.0\,\text{L}\) of \(0.6\,\text{M HCl}\)?

Answer: 100 mL

Explanation:
$$M_1 V_1 = M_2 V_2$$ $$12 \times V_1 = 0.6 \times 2000 \implies 12 V_1 = 1200 \implies V_1 = 100\,\text{mL}$$

118. Which sample contains the maximum number of atoms? (A) \(2\,\text{g H}_2\) (B) \(16\,\text{g O}_2\) (C) \(28\,\text{g N}_2\) (D) \(18\,\text{g H}_2\text{O}\)

Answer: 18 g H2O

Explanation:
Atoms in \(2\,\text{g H}_2 = 1 \times 2 \times N_A = 2 N_A\).
Atoms in \(16\,\text{g O}_2 = 0.5 \times 2 \times N_A = 1 N_A\).
Atoms in \(28\,\text{g N}_2 = 1 \times 2 \times N_A = 2 N_A\).
Atoms in \(18\,\text{g H}_2\text{O} = 1 \times 3 \times N_A = 3 N_A\) (Maximum).

119. State Gay-Lussac's Law of Combining Volumes.

Answer: Gases combine in simple integer volume ratios

Explanation:
Gay-Lussac's law states that when gases react together, they do so in volumes which bear a simple whole-number ratio to one another and to the volumes of gaseous products, provided temperature and pressure remain constant.

120. Calculate the mass of sodium carbonate (\(\text{Na}_2\text{CO}_3\)) required to react with \(100\,\text{mL}\) of \(0.5\,\text{M H}_2\text{SO}_4\). [Molar mass = \(106\,\text{g/mol}\)]

Answer: 5.3 g

Explanation:
$$\text{Na}_2\text{CO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} + \text{CO}_2$$ Moles of \(\text{H}_2\text{SO}_4 = 0.5 \times 0.1 = 0.05\,\text{mol}\).
Moles of \(\text{Na}_2\text{CO}_3\) required \(= 0.05\,\text{mol}\).
$$\text{Mass} = 0.05 \times 106 = 5.3\,\text{g}$$

121. What is the mass of \(1\,\text{L}\) of \(\text{O}_2\) gas at STP?

Answer: 1.43 g

Explanation:
$$\text{Density at STP} = \frac{32\,\text{g}}{22.4\,\text{L}} \approx 1.43\,\text{g/L}$$ Mass of \(1\,\text{L} = 1.43\,\text{g}\).

122. Find the total number of ions present in \(11.1\,\text{g}\) of \(\text{CaCl}_2\). [Molar mass of \(\text{CaCl}_2 = 111\,\text{g/mol}\)]

Answer: 1.8066 \times 10^{23}

Explanation:
Moles of \(\text{CaCl}_2 = \frac{11.1}{111} = 0.1\,\text{mol}\).
\(1\,\text{formula unit of CaCl}_2\) dissociates into 3 ions (\(1\text{Ca}^{2+} + 2\text{Cl}^-\)).
$$\text{Total ions} = 0.1 \times 3 \times 6.022 \times 10^{23} = 1.8066 \times 10^{23}$$

123. Calculate the empirical formula mass of ethanoic acid (\(\text{CH}_3\text{COOH}\)).

Answer: 30 u

Explanation:
Molecular formula \(= \text{C}_2\text{H}_4\text{O}_2\). Empirical formula \(= \text{CH}_2\text{O}\).
$$\text{Empirical formula mass} = 12 + 2(1) + 16 = 30\,\text{u}$$

124. What volume of \(0.1\,\text{M KMnO}_4\) in acidic medium is required to oxidize \(20\,\text{mL}\) of \(0.5\,\text{M FeSO}_4\)?

Answer: 20 mL

Explanation:
$$\text{Milli-equivalents of KMnO}_4 = \text{Milli-equivalents of FeSO}_4$$ $$V_1 \times 0.1 \times 5 = 20 \times 0.5 \times 1$$ $$0.5 V_1 = 10 \implies V_1 = 20\,\text{mL}$$

125. Convert \(500\,\text{mL}\) to \(\text{m}^3\).

Answer: 5 \times 10^{-4} m^3

Explanation:
$$1\,\text{mL} = 10^{-6}\,\text{m}^3$$ $$500\,\text{mL} = 500 \times 10^{-6} = 5 \times 10^{-4}\,\text{m}^3$$

126. How many moles of \(\text{CO}_2\) are produced when \(2\,\text{moles}\) of butane (\(\text{C}_4\text{H}_{10}\)) undergo complete combustion?

Answer: 8 moles

Explanation:
$$2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2\text{O}$$ \(2\,\text{moles of C}_4\text{H}_{10}\) directly produce \(8\,\text{moles of CO}_2\).

127. Calculate the mass of glucose needed to prepare \(250\,\text{mL}\) of a \(5\%\) (w/v) solution.

Answer: 12.5 g

Explanation:
\(5\%\) (w/v) means \(5\,\text{g}\) in \(100\,\text{mL}\) solution.
$$\text{Mass in 250 mL} = \frac{5}{100} \times 250 = 12.5\,\text{g}$$

128. What is the number of moles of neutrons in \(36\,\text{g}\) of heavy water (\(\text{D}_2\text{O}\))? [Atomic mass of \(\text{D} = 2\)]

Answer: 18 moles

Explanation:
Molar mass of \(\text{D}_2\text{O} = (2 \times 2) + 16 = 20\,\text{g/mol}\).
Moles of \(\text{D}_2\text{O} = \frac{36}{20} = 1.8\,\text{mol}\).
Neutrons in 1 molecule of \(\text{D}_2\text{O} = 1(\text{D}) + 1(\text{D}) + 8(\text{O}) = 10\).
$$\text{Moles of neutrons} = 1.8 \times 10 = 18\,\text{moles}$$

129. Round off \(0.007895\) to 3 significant figures.

Answer: 0.00790

Explanation:
The leading zeros are non-significant. The 4th significant digit is 5, rounding 9 up yields \(0.00790\).

130. A gaseous mixture contains \(2\,\text{g}\) of \(\text{H}_2\) and \(8\,\text{g}\) of \(\text{He}\). What is the mole fraction of \(\text{H}_2\)?

Answer: 0.333

Explanation:
Moles of \(\text{H}_2 = \frac{2}{2} = 1\,\text{mol}\).
Moles of \(\text{He} = \frac{8}{4} = 2\,\text{mol}\).
$$\text{Mole fraction of H}_2 = \frac{1}{1 + 2} = \frac{1}{3} \approx 0.333$$

131. Calculate the density of nitrogen gas (\(\text{N}_2\)) at \(27^\circ\text{C}\) and \(1\,\text{atm}\) pressure. [\(\text{R} = 0.0821\,\text{L atm K}^{-1}\text{mol}^{-1}\)]

Answer: 1.137 g/L

Explanation:
$$P M = d R T \implies d = \frac{P M}{R T}$$ $$d = \frac{1 \times 28}{0.0821 \times 300} = \frac{28}{24.63} \approx 1.137\,\text{g/L}$$

132. What mass of \(\text{KClO}_3\) is required to produce \(6.72\,\text{L}\) of oxygen gas at STP? [Molar mass of \(\text{KClO}_3 = 122.5\,\text{g/mol}\)]

Answer: 24.5 g

Explanation:
Moles of \(\text{O}_2 = \frac{6.72}{22.4} = 0.3\,\text{mol}\).
\(2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2\).
Moles of \(\text{KClO}_3\) required \(= 0.3 \times \frac{2}{3} = 0.2\,\text{mol}\).
$$\text{Mass} = 0.2 \times 122.5 = 24.5\,\text{g}$$

133. Calculate the molarity of a solution made by dissolving \(4\,\text{g}\) of \(\text{NaOH}\) in enough water to make \(250\,\text{mL}\) of solution.

Answer: 0.4 M

Explanation:
$$\text{Moles of NaOH} = \frac{4}{40} = 0.1\,\text{mol}$$ $$\text{Molarity} = \frac{0.1}{0.25} = 0.4\,\text{M}$$

134. What is the equivalent mass of oxalic acid dihydrate (\(\text{H}_2\text{C}_2\text{O}_4\cdot 2\text{H}_2\text{O}\))? [Molar mass = \(126\,\text{g/mol}\)]

Answer: 63 g/equiv

Explanation:
Oxalic acid is a dibasic acid (\(n\text{-factor} = 2\)).
$$\text{Equivalent Mass} = \frac{126}{2} = 63\,\text{g/equiv}$$

135. Find the total number of atoms in \(0.1\,\text{mole}\) of \(\text{P}_4\).

Answer: 2.4088 \times 10^{23}

Explanation:
Moles of P atoms \(= 0.1 \times 4 = 0.4\,\text{mol}\).
$$\text{Number of atoms} = 0.4 \times 6.022 \times 10^{23} = 2.4088 \times 10^{23}$$

136. What is the mole fraction of solute in a binary solution if the molality is \(2.0\,\text{m}\) and the solvent is water?

Answer: 0.0347

Explanation:
Moles of solute \(= 2.0\,\text{mol}\). Moles of water \(= \frac{1000}{18} = 55.55\,\text{mol}\).
$$\chi_{\text{solute}} = \frac{2.0}{2.0 + 55.55} = \frac{2.0}{57.55} \approx 0.0347$$

137. How many grams of \(\text{CaCl}_2\) are required to prepare \(500\,\text{mL}\) of a \(0.2\,\text{M}\) solution? [Molar mass = \(111\,\text{g/mol}\)]

Answer: 11.1 g

Explanation:
$$\text{Moles} = 0.2 \times 0.5 = 0.1\,\text{mol}$$ $$\text{Mass} = 0.1 \times 111 = 11.1\,\text{g}$$

138. How many moles of \(\text{O}_2\) are needed to produce \(4.4\,\text{g}\) of \(\text{CO}_2\) from combustion of carbon?

Answer: 0.1 mole

Explanation:
$$\text{C} + \text{O}_2 \rightarrow \text{CO}_2$$ Moles of \(\text{CO}_2 = \frac{4.4}{44} = 0.1\,\text{mol}\).
Moles of \(\text{O}_2\) required \(= 0.1\,\text{mol}\).

139. State the formula relating molarity (\(M\)), density of solution (\(d\) in g/mL), mass percent (\(W\%\)), and solute molar mass (\(M_B\)).

Answer: M = (10 * W% * d) / M_B

Explanation:
$$\text{Molarity} (M) = \frac{10 \times (W\%) \times d}{M_B}$$

140. What volume of carbon dioxide at STP is produced by the complete decomposition of \(5\,\text{g}\) of \(\text{CaCO}_3\)?

Answer: 1.12 L

Explanation:
Moles of \(\text{CaCO}_3 = \frac{5}{100} = 0.05\,\text{mol}\).
Moles of \(\text{CO}_2 = 0.05\,\text{mol}\).
$$\text{Volume at STP} = 0.05 \times 22.4 = 1.12\,\text{L}$$

141. Determine the mass of nitrogen present in \(0.5\,\text{moles}\) of ammonium nitrate (\(\text{NH}_4\text{NO}_3\)).

Answer: 14 g

Explanation:
\(1\,\text{mole of NH}_4\text{NO}_3\) contains \(2\,\text{moles of N atoms}\) (\(28\,\text{g}\)).
$$\text{Mass of N in 0.5 mol} = 0.5 \times 28 = 14\,\text{g}$$

142. If law of conservation of mass holds true, how much \(\text{CaO}\) is formed when \(50\,\text{g}\) of \(\text{CaCO}_3\) decomposes to give \(22\,\text{g}\) of \(\text{CO}_2\)?

Answer: 28 g

Explanation:
$$\text{Mass of reactants} = \text{Mass of products}$$ $$50\,\text{g} = \text{Mass of CaO} + 22\,\text{g} \implies \text{Mass of CaO} = 28\,\text{g}$$

143. Express \(0.00078\) in standard scientific notation.

Answer: 7.8 \times 10^{-4}

Explanation:
Moving the decimal point 4 positions to the right yields \(7.8 \times 10^{-4}\).

144. How many water molecules are present in a crystal of copper sulfate weighing \(24.95\,\text{g}\) (\(\text{CuSO}_4\cdot 5\text{H}_2\text{O}\))? [Molar mass = \(249.5\,\text{g/mol}\)]

Answer: 3.011 \times 10^{23}

Explanation:
Moles of compound \(= \frac{24.95}{249.5} = 0.1\,\text{mol}\).
Moles of water molecules \(= 0.1 \times 5 = 0.5\,\text{mol}\).
$$\text{Number of water molecules} = 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}$$

145. Calculate the normality of a solution prepared by dissolving \(4.9\,\text{g}\) of \(\text{H}_2\text{SO}_4\) in \(500\,\text{mL}\) of solution.

Answer: 0.2 N

Explanation:
$$\text{Equivalent mass of H}_2\text{SO}_4 = \frac{98}{2} = 49\,\text{g/equiv}$$ $$\text{Number of equivalents} = \frac{4.9}{49} = 0.1$$ $$\text{Normality} = \frac{0.1}{0.5\,\text{L}} = 0.2\,\text{N}$$

146. What is the mass of \(10^{21}\) molecules of oxygen gas (\(\text{O}_2\))?

Answer: 0.053 g

Explanation:
$$\text{Moles} = \frac{10^{21}}{6.022 \times 10^{23}} \approx 0.00166\,\text{mol}$$ $$\text{Mass} = 0.00166 \times 32 \approx 0.053\,\text{g}$$

147. A gas has an empirical formula \(\text{CH}_2\) and its molar mass is \(56\,\text{g/mol}\). What is its molecular formula?

Answer: C4H8

Explanation:
$$\text{Empirical formula mass} = 12 + 2 = 14\,\text{u}$$ $$n = \frac{\text{Molecular mass}}{\text{Empirical mass}} = \frac{56}{14} = 4$$ $$\text{Molecular formula} = (\text{CH}_2)_4 = \text{C}_4\text{H}_8$$

148. Calculate the mole fraction of \(\text{NaOH}\) in a solution containing \(20\,\text{g}\) \(\text{NaOH}\) and \(72\,\text{g}\) water.

Answer: 0.111

Explanation:
Moles of \(\text{NaOH} = \frac{20}{40} = 0.5\,\text{mol}\).
Moles of water \(= \frac{72}{18} = 4.0\,\text{mol}\).
$$\chi_{\text{NaOH}} = \frac{0.5}{0.5 + 4.0} = \frac{0.5}{4.5} = \frac{1}{9} \approx 0.111$$

149. What volume of \(0.5\,\text{M HCl}\) is neutralized by \(100\,\text{mL}\) of \(0.25\,\text{M Ca(OH)}_2\)?

Answer: 100 mL

Explanation:
$$\text{Milli-equivalents of Ca(OH)}_2 = 100 \times 0.25 \times 2 = 50$$ $$\text{Milli-equivalents of HCl} = V \times 0.5 \times 1 = 0.5 V$$ $$0.5 V = 50 \implies V = 100\,\text{mL}$$

150. State the total number of electrons in \(1\,\text{mole}\) of hydrogen gas (\(\text{H}_2\)).

Answer: 1.2044 \times 10^{24}

Explanation:
Each molecule of \(\text{H}_2\) has 2 electrons.
$$\text{Total electrons} = 2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{24}$$

151. What is the total number of subatomic particles (protons, neutrons, electrons) present in \(0.1\,\text{mole}\) of \(\text{CO}_2\)?

Answer: 3.9745 \times 10^{24}

Explanation:
A molecule of \(\text{CO}_2\) has:
Protons \(= 6 + 8 + 8 = 22\), Electrons \(= 22\), Neutrons \(= 6 + 8 + 8 = 22\).
Total particles per molecule \(= 66\).
$$\text{Total particles} = 0.1 \times 66 \times 6.022 \times 10^{23} = 3.9745 \times 10^{24}$$

152. Calculate the volume of \(0.2\,\text{M H}_2\text{SO}_4\) needed to completely neutralize \(50\,\text{mL}\) of \(0.4\,\text{M KOH}\).

Answer: 50 mL

Explanation:
$$\text{Milli-equivalents of acid} = \text{Milli-equivalents of base}$$ $$V_1 \times 0.2 \times 2 = 50 \times 0.4 \times 1$$ $$0.4 V_1 = 20 \implies V_1 = 50\,\text{mL}$$

153. What is the concentration in ppm (parts per million) if \(5\,\text{mg}\) of \(\text{CaCO}_3\) is present in \(1\,\text{kg}\) of water?

Answer: 5 ppm

Explanation:
$$\text{ppm} = \frac{\text{Mass of solute (g)}}{\text{Mass of solution (g)}} \times 10^6 = \frac{5 \times 10^{-3}\,\text{g}}{1000\,\text{g}} \times 10^6 = 5\,\text{ppm}$$

154. How many grams of pure oxygen are needed to completely burn \(16\,\text{g}\) of methane (\(\text{CH}_4\))?

Answer: 64 g

Explanation:
$$\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$$ \(16\,\text{g CH}_4 = 1\,\text{mole}\).
Requires \(2\,\text{moles of O}_2 = 2 \times 32 = 64\,\text{g}\).

155. Calculate the mass of one atom of nitrogen in grams.

Answer: 2.325 \times 10^{-23} g

Explanation:
$$\text{Mass} = \frac{\text{Atomic mass}}{N_A} = \frac{14}{6.022 \times 10^{23}} \approx 2.325 \times 10^{-23}\,\text{g}$$

156. If the atomic mass of carbon was redefined as 6 u instead of 12 u, what would be the mass of 1 mole of any substance?

Answer: Half of its present mass

Explanation:
Redefining the atomic mass unit scale by a factor of 0.5 halves the mass of 1 mole of any given substance relative to standard standard units.

157. What is the oxidation state of sulfur in sodium thiosulfate (\(\text{Na}_2\text{S}_2\text{O}_3\))?

Answer: +2

Explanation:
$$2(+1) + 2(x) + 3(-2) = 0 \implies 2x - 4 = 0 \implies x = +2$$

158. Find the percentage composition of carbon in ethanol (\(\text{C}_2\text{H}_5\text{OH}\)). [Molar mass = \(46\,\text{g/mol}\)]

Answer: 52.17%

Explanation:
$$\text{Mass of Carbon} = 2 \times 12 = 24\,\text{g}$$ $$\%\text{ Carbon} = \frac{24}{46} \times 100 \approx 52.17\%$$

159. What is the number of moles of water formed when \(4\,\text{g}\) of hydrogen gas reacts with \(16\,\text{g}\) of oxygen gas?

Answer: 1.0 mole

Explanation:
$$2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$$ Moles of \(\text{H}_2 = 2\,\text{mol}\), Moles of \(\text{O}_2 = \frac{16}{32} = 0.5\,\text{mol}\).
\(\text{O}_2\) is the limiting reagent.
\(0.5\,\text{mole O}_2\) produces \(2 \times 0.5 = 1.0\,\text{mole H}_2\text{O}\).

160. Calculate the number of moles of \(\text{Al}^{3+}\) ions in \(51\,\text{g}\) of \(\text{Al}_2\text{O}_3\). [Molar mass of \(\text{Al}_2\text{O}_3 = 102\,\text{g/mol}\)]

Answer: 1.0 mole

Explanation:
Moles of \(\text{Al}_2\text{O}_3 = \frac{51}{102} = 0.5\,\text{mol}\).
Each mole of \(\text{Al}_2\text{O}_3\) contains \(2\,\text{moles of Al}^{3+}\) ions.
$$\text{Moles of Al}^{3+} = 0.5 \times 2 = 1.0\,\text{mole}$$

161. Express the product of \(4.5 \times 10^3\) and \(2.0 \times 10^4\) in scientific notation.

Answer: 9.0 \times 10^7

Explanation:
$$(4.5 \times 2.0) \times 10^{3+4} = 9.0 \times 10^7$$

162. How many moles of \(\text{HCl}\) are present in \(40\,\text{mL}\) of a \(0.25\,\text{M}\) solution?

Answer: 0.01 moles

Explanation:
$$\text{Moles} = \text{Molarity} \times \text{Volume in L} = 0.25 \times 0.04 = 0.01\,\text{mol}$$

163. Calculate the percentage of hydrated water in washing soda (\(\text{Na}_2\text{CO}_3\cdot 10\text{H}_2\text{O}\)). [Molar mass = \(286\,\text{g/mol}\)]

Answer: 62.94%

Explanation:
$$\text{Mass of 10H}_2\text{O} = 10 \times 18 = 180\,\text{g}$$ $$\%\text{ Water} = \frac{180}{286} \times 100 \approx 62.94\%$$

164. What is the equivalent mass of \(\text{H}_3\text{PO}_3\) (phosphorous acid) when it acts as a dibasic acid? [Molar mass = \(M\)]

Answer: M / 2

Explanation:
\(\text{H}_3\text{PO}_3\) has 2 replaceable hydrogen atoms (\(n\text{-factor} = 2\)).
$$\text{Equivalent Mass} = \frac{M}{2}$$

165. Determine the volume of \(0.1\,\text{M NaOH}\) needed to neutralize \(25\,\text{mL}\) of \(0.1\,\text{M H}_2\text{SO}_4\).

Answer: 50 mL

Explanation:
$$\text{Milli-equivalents of acid} = 25 \times 0.1 \times 2 = 5$$ $$\text{Milli-equivalents of base} = V \times 0.1 \times 1 = 0.1 V$$ $$0.1 V = 5 \implies V = 50\,\text{mL}$$

166. How many molecules of nitrogen are present in a container of volume \(1.12\,\text{L}\) at STP?

Answer: 3.011 \times 10^{22}

Explanation:
$$\text{Moles} = \frac{1.12}{22.4} = 0.05\,\text{mol}$$ $$\text{Molecules} = 0.05 \times 6.022 \times 10^{23} = 3.011 \times 10^{22}$$

167. A compound contains \(40\%\) carbon, \(6.67\%\) hydrogen, and \(53.33\%\) oxygen. What is its empirical formula?

Answer: CH2O

Explanation:
Moles of \(\text{C} = \frac{40}{12} = 3.33\); Moles of \(\text{H} = \frac{6.67}{1} = 6.67\); Moles of \(\text{O} = \frac{53.33}{16} = 3.33\).
Ratio \(\text{C:H:O} = 1 : 2 : 1\). Empirical formula is \(\text{CH}_2\text{O}\).

168. What volume of carbon monoxide at STP is required to reduce \(16\,\text{g}\) of \(\text{Fe}_2\text{O}_3\) completely to iron? [Molar mass of \(\text{Fe}_2\text{O}_3 = 160\,\text{g/mol}\)]

Answer: 6.72 L

Explanation:
$$\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$$ Moles of \(\text{Fe}_2\text{O}_3 = \frac{16}{160} = 0.1\,\text{mol}\).
Moles of \(\text{CO}\) needed \(= 3 \times 0.1 = 0.3\,\text{mol}\).
$$\text{Volume at STP} = 0.3 \times 22.4 = 6.72\,\text{L}$$

169. What is the molality of pure water at \(4^\circ\text{C}\) (density \(= 1.0\,\text{g/mL}\))?

Answer: 55.55 m

Explanation:
Consider \(1\,\text{kg}\) of water (\(1000\,\text{g}\)).
$$\text{Moles of solute (water)} = \frac{1000}{18} = 55.55\,\text{mol}$$ $$\text{Molality} = \frac{55.55\,\text{mol}}{1\,\text{kg}} = 55.55\,\text{m}$$

170. How many electrons are lost when \(1\,\text{mole}\) of \(\text{Fe}^{2+}\) is oxidized to \(\text{Fe}^{3+}\)?

Answer: 6.022 \times 10^{23}

Explanation:
$$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$$ \(1\,\text{mole of Fe}^{2+}\) releases \(1\,\text{mole of electrons} = 6.022 \times 10^{23}\) electrons.

171. Round off \(12.450\) to 3 significant figures using standard rounding rules.

Answer: 12.4

Explanation:
When rounding off to 3 significant figures, the digit after 4 is 5 followed by 0. By the odd-even rule, the preceding digit (4) is even, so it remains unchanged: \(12.4\).

172. Calculate the mass of \(\text{Na}_2\text{SO}_4\) required to prepare \(100\,\text{mL}\) of a \(0.5\,\text{M}\) solution. [Molar mass = \(142\,\text{g/mol}\)]

Answer: 7.1 g

Explanation:
$$\text{Moles} = 0.5 \times 0.1 = 0.05\,\text{mol}$$ $$\text{Mass} = 0.05 \times 142 = 7.1\,\text{g}$$

173. What is the total number of atoms in \(1\,\text{mole}\) of \(\text{K}_4[\text{Fe(CN)}_6]\)?

Answer: 1.0237 \times 10^{25}

Explanation:
Atoms per formula unit \(= 4 (\text{K}) + 1 (\text{Fe}) + 6 (\text{C}) + 6 (\text{N}) = 17\).
$$\text{Total atoms} = 17 \times 6.022 \times 10^{23} \approx 1.0237 \times 10^{25}$$

174. If \(30\,\text{mL}\) of \(0.5\,\text{M H}_2\text{SO}_4\) is mixed with \(20\,\text{mL}\) of \(1.0\,\text{M H}_2\text{SO}_4\), what is the final molarity?

Answer: 0.7 M

Explanation:
$$M_{\text{mix}} = \frac{M_1 V_1 + M_2 V_2}{V_1 + V_2} = \frac{(0.5 \times 30) + (1.0 \times 20)}{30 + 20} = \frac{15 + 20}{50} = 0.7\,\text{M}$$

175. What mass of \(\text{CO}_2\) is produced when \(1\,\text{mole}\) of glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)) undergoes complete respiration?

Answer: 264 g

Explanation:
$$\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O}$$ \(1\,\text{mole glucose}\) produces \(6\,\text{moles CO}_2\).
$$\text{Mass of CO}_2 = 6 \times 44 = 264\,\text{g}$$

176. What is the law illustrated by the formation of \(\text{CO}\) and \(\text{CO}_2\) from carbon and oxygen?

Answer: Law of Multiple Proportions

Explanation:
When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers.

177. How many moles of sulfur atoms are in \(64\,\text{g}\) of \(\text{S}_8\)? [Atomic mass of \(\text{S} = 32\)]

Answer: 2.0 moles

Explanation:
Total mass of sulfur \(= 64\,\text{g}\).
$$\text{Moles of S atoms} = \frac{64}{32} = 2.0\,\text{moles}$$

178. Calculate the volume occupied by \(4.4\,\text{g}\) of \(\text{CO}_2\) at \(27^\circ\text{C}\) and \(1\,\text{atm}\) pressure. [\(\text{R} = 0.0821\,\text{L atm K}^{-1}\text{mol}^{-1}\)]

Answer: 2.46 L

Explanation:
$$\text{Moles} = \frac{4.4}{44} = 0.1\,\text{mol}$$ $$V = \frac{n R T}{P} = \frac{0.1 \times 0.0821 \times 300}{1} = 2.463\,\text{L}$$

179. Find the mass percentage of chlorine in sodium chloride (\(\text{NaCl}\)). [Atomic mass: \(\text{Na}=23\), \(\text{Cl}=35.5\)]

Answer: 60.68%

Explanation:
$$\text{Molar mass of NaCl} = 58.5\,\text{g/mol}$$ $$\%\text{ Chlorine} = \frac{35.5}{58.5} \times 100 \approx 60.68\%$$

180. How many unpaired electrons are present in a neutral nitrogen atom (\(Z = 7\))?

Answer: 3

Explanation:
Electronic configuration of \(\text{N}\) is \(1s^2 2s^2 2p^3\). According to Hund's rule, all 3 electrons in the \(2p\) subshell are unpaired.

181. Calculate the mole fraction of water in a solution prepared by dissolving \(18\,\text{g}\) of glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)) in \(162\,\text{g}\) of water.

Answer: 0.989

Explanation:
Moles of glucose \(= \frac{18}{180} = 0.1\,\text{mol}\).
Moles of water \(= \frac{162}{18} = 9.0\,\text{mol}\).
$$\chi_{\text{water}} = \frac{9.0}{9.0 + 0.1} = \frac{9.0}{9.1} \approx 0.989$$

182. What is the equivalent mass of \(\text{Al(OH)}_3\)? [Molar mass = \(78\,\text{g/mol}\)]

Answer: 26 g/equiv

Explanation:
\(\text{Al(OH)}_3\) has 3 replaceable \(\text{OH}^-\) groups (\(n\text{-factor} = 3\)).
$$\text{Equivalent Mass} = \frac{78}{3} = 26\,\text{g/equiv}$$

183. How many liters of \(\text{H}_2\) gas at STP are needed to react with \(11.2\,\text{L}\) of \(\text{O}_2\) gas to form water?

Answer: 22.4 L

Explanation:
$$2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$$ \(1\,\text{volume O}_2\) requires \(2\,\text{volumes H}_2\).
$$\text{Volume of H}_2 = 2 \times 11.2 = 22.4\,\text{L}$$

184. What is the mass of \(0.5\,\text{moles}\) of ozone (\(\text{O}_3\)) gas?

Answer: 24 g

Explanation:
$$\text{Molar mass of O}_3 = 3 \times 16 = 48\,\text{g/mol}$$ $$\text{Mass} = 0.5 \times 48 = 24\,\text{g}$$

185. State the law of reciprocal proportions formulator.

Answer: Jeremias Richter

Explanation:
The Law of Reciprocal Proportions was proposed by Jeremias Richter in 1792.

186. Determine the volume of \(0.2\,\text{M HCl}\) required to react completely with \(1.06\,\text{g}\) of anhydrous \(\text{Na}_2\text{CO}_3\). [Molar mass = \(106\,\text{g/mol}\)]

Answer: 100 mL

Explanation:
$$\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$$ Moles of \(\text{Na}_2\text{CO}_3 = \frac{1.06}{106} = 0.01\,\text{mol}\).
Moles of \(\text{HCl}\) required \(= 0.02\,\text{mol}\).
$$V = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.02}{0.2} = 0.1\,\text{L} = 100\,\text{mL}$$

187. How many moles of electrons are required to discharge \(1\,\text{mole}\) of \(\text{Cr}^{3+}\) to \(\text{Cr}\) metal?

Answer: 3 moles

Explanation:
$$\text{Cr}^{3+} + 3e^- \rightarrow \text{Cr}$$ \(1\,\text{mole of Cr}^{3+}\) requires \(3\,\text{moles of electrons}\).

188. Calculate the molarity of a solution containing \(10\,\text{g}\) of \(\text{KOH}\) dissolved in \(500\,\text{mL}\) solution. [Molar mass of \(\text{KOH} = 56\,\text{g/mol}\)]

Answer: 0.357 M

Explanation:
$$\text{Moles of KOH} = \frac{10}{56} \approx 0.1786\,\text{mol}$$ $$\text{Molarity} = \frac{0.1786}{0.5\,\text{L}} \approx 0.357\,\text{M}$$

189. Find the mass of oxygen present in \(1\,\text{mole}\) of potassium dichromate (\(\text{K}_2\text{Cr}_2\text{O}_7\)).

Answer: 112 g

Explanation:
\(1\,\text{mole of K}_2\text{Cr}_2\text{O}_7\) contains \(7\,\text{moles of O atoms}\).
$$\text{Mass of O} = 7 \times 16 = 112\,\text{g}$$

190. What is the ratio of rates of diffusion of \(\text{H}_2\) gas and \(\text{O}_2\) gas under identical conditions?

Answer: 4 : 1

Explanation:
According to Graham's law of diffusion: $$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4 : 1$$

191. How many total atoms are present in \(0.5\,\text{mole}\) of sucrose (\(\text{C}_{12}\text{H}_{22}\text{O}_{11}\))?

Answer: 1.355 \times 10^{25}

Explanation:
Atoms per molecule \(= 12 + 22 + 11 = 45\).
$$\text{Total atoms} = 0.5 \times 45 \times 6.022 \times 10^{23} \approx 1.355 \times 10^{25}$$

192. What volume of water must be added to \(500\,\text{mL}\) of \(1.0\,\text{M NaOH}\) to make its concentration \(0.2\,\text{M}\)?

Answer: 2000 mL

Explanation:
$$M_1 V_1 = M_2 V_2 \implies 1.0 \times 500 = 0.2 \times V_2 \implies V_2 = 2500\,\text{mL}$$ $$\text{Water added} = 2500 - 500 = 2000\,\text{mL}$$

193. Calculate the percentage of calcium in calcium phosphate (\(\text{Ca}_3(\text{PO}_4)_2\)). [Molar mass = \(310\,\text{g/mol}\), \(\text{Ca}=40\)]

Answer: 38.71%

Explanation:
$$\text{Mass of Ca} = 3 \times 40 = 120\,\text{g}$$ $$\%\text{ Calcium} = \frac{120}{310} \times 100 \approx 38.71\%$$

194. Round off \(0.003405\) to 3 significant figures.

Answer: 0.00341

Explanation:
Leading zeros are non-significant. The 4th significant digit is 5, rounding up the preceding digit gives \(0.00341\).

195. How many moles of \(\text{KMnO}_4\) are required to completely oxidize \(1\,\text{mole}\) of oxalate ions (\(\text{C}_2\text{O}_4^{2-}\)) in acidic medium?

Answer: 0.4 moles

Explanation:
$$\text{Equivalents of KMnO}_4 = \text{Equivalents of C}_2\text{O}_4^{2-}$$ \(n\)-factor of \(\text{KMnO}_4 = 5\); \(n\)-factor of \(\text{C}_2\text{O}_4^{2-} = 2\).
$$n \times 5 = 1 \times 2 \implies n = 0.4\,\text{moles}$$

196. Find the empirical formula of a compound containing \(75\%\) carbon and \(25\%\) hydrogen by mass.

Answer: CH4

Explanation:
Moles of \(\text{C} = \frac{75}{12} = 6.25\); Moles of \(\text{H} = \frac{25}{1} = 25\).
Ratio \(\text{C:H} = 1 : 4\). Empirical formula is \(\text{CH}_4\).

197. Calculate the mass of \(\text{BaSO}_4\) precipitated when excess \(\text{BaCl}_2\) solution is added to \(100\,\text{mL}\) of \(0.1\,\text{M H}_2\text{SO}_4\). [Molar mass of \(\text{BaSO}_4 = 233\,\text{g/mol}\)]

Answer: 2.33 g

Explanation:
Moles of \(\text{H}_2\text{SO}_4 = 0.1 \times 0.1 = 0.01\,\text{mol}\).
Moles of \(\text{BaSO}_4\) formed \(= 0.01\,\text{mol}\).
$$\text{Mass of BaSO}_4 = 0.01 \times 233 = 2.33\,\text{g}$$

198. What is the density of gaseous \(\text{CO}_2\) at STP?

Answer: 1.96 g/L

Explanation:
$$\text{Density} = \frac{\text{Molar mass}}{\text{Molar volume at STP}} = \frac{44\,\text{g}}{22.4\,\text{L}} \approx 1.96\,\text{g/L}$$

199. How many moles of \(\text{NH}_3\) can be produced from \(2\,\text{moles}\) of \(\text{N}_2\) and \(3\,\text{moles}\) of \(\text{H}_2\)?

Answer: 2 moles

Explanation:
$$\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$$ \(3\,\text{moles of H}_2\) react with \(1\,\text{mole of N}_2\) to produce \(2\,\text{moles of NH}_3\).
\(\text{H}_2\) is the limiting reagent and directly yields \(2\,\text{moles of NH}_3\).

200. Calculate the number of chloride ions present in \(5.85\,\text{g}\) of sodium chloride (\(\text{NaCl}\)).

Answer: 6.022 \times 10^{22}

Explanation:
$$\text{Moles of NaCl} = \frac{5.85}{58.5} = 0.1\,\text{mol}$$ $$\text{Moles of Cl}^- = 0.1\,\text{mol}$$ $$\text{Number of Cl}^- \text{ ions} = 0.1 \times 6.022 \times 10^{23} = 6.022 \times 10^{22}$$

201. What is the wavelength of a photon having an energy of \(3.03 \times 10^{-19}\,\text{J}\)? [\(h = 6.626 \times 10^{-34}\,\text{J s}\), \(c = 3 \times 10^8\,\text{m/s}\)]

Answer: 656 nm

Explanation:
$$\lambda = \frac{h c}{E} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{3.03 \times 10^{-19}} \approx 6.56 \times 10^{-7}\,\text{m} = 656\,\text{nm}$$

202. How many subshells are present in the principal energy level \(n = 4\)?

Answer: 4

Explanation:
For a principal quantum number \(n\), the number of subshells is equal to \(n\). For \(n=4\), the subshells are \(4s, 4p, 4d,\) and \(4f\).

203. What is the maximum number of electrons that can be accommodated in the \(3d\) subshell?

Answer: 10

Explanation:
The \(d\) subshell contains 5 orbitals (\(l = 2\)). Each orbital can hold a maximum of 2 electrons, giving \(5 \times 2 = 10\) electrons.

204. Calculate the frequency of radiation having a wavelength of \(600\,\text{nm}\).

Answer: 5.0 \times 10^{14} Hz

Explanation:
$$\nu = \frac{c}{\lambda} = \frac{3 \times 10^8\,\text{m/s}}{600 \times 10^{-9}\,\text{m}} = 5.0 \times 10^{14}\,\text{Hz}$$

205. What is the de Broglie wavelength associated with an electron moving with a velocity of \(10^6\,\text{m/s}\)? [\(m_e = 9.1 \times 10^{-31}\,\text{kg}\)]

Answer: 7.28 \times 10^{-10} m

Explanation:
$$\lambda = \frac{h}{m v} = \frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 10^6} \approx 7.28 \times 10^{-10}\,\text{m}$$

206. Which quantum number determines the spatial orientation of an orbital?

Answer: Magnetic quantum number (m_l)

Explanation:
The magnetic quantum number (\(m_l\)) describes the orientation of the orbital in space relative to a coordinate axis.

207. What is the radius of the first Bohr orbit of a hydrogen atom?

Answer: 0.529 Å (52.9 pm)

Explanation:
The radius formula is \(r_n = 0.529 \times \frac{n^2}{Z}\,\text{Å}\). For \(n=1, Z=1\), \(r_1 = 0.529\,\text{Å}\).

208. Calculate the energy of an electron in the ground state of a hydrogen atom in electron-volts (\(\text{eV}\)).

Answer: -13.6 eV

Explanation:
$$E_n = -13.6 \times \frac{Z^2}{n^2}\,\text{eV}$$ For ground state (\(n=1, Z=1\)), \(E_1 = -13.6\,\text{eV}\).

209. How many spherical (radial) nodes are present in a \(4s\) orbital?

Answer: 3

Explanation:
$$\text{Radial nodes} = n - l - 1$$ For \(4s\): \(n = 4, l = 0\). Radial nodes \(= 4 - 0 - 1 = 3\).

210. What is the ground state electronic configuration of Chromium (\(Z = 24\))?

Answer: [Ar] 3d^5 4s^1

Explanation:
Chromium exhibits an anomalous configuration due to the extra stability of half-filled \(d\)-subshells, transferring an electron from \(4s\) to \(3d\).

211. Calculate the energy required to ionize a hydrogen atom from its ground state.

Answer: 13.6 eV (or 2.18 \times 10^{-18} J)

Explanation:
$$\text{Ionization Energy} = E_\infty - E_1 = 0 - (-13.6\,\text{eV}) = 13.6\,\text{eV}$$

212. What is the total number of orbitals associated with the principal quantum number \(n = 3\)?

Answer: 9

Explanation:
Total orbitals \(= n^2 = 3^2 = 9\) (one \(3s\), three \(3p\), five \(3d\)).

213. State the total orbital angular momentum of an electron in an \(s\)-orbital.

Answer: 0

Explanation:
$$\text{Orbital angular momentum} = \sqrt{l(l+1)} \frac{h}{2\pi}$$ For an \(s\)-orbital, \(l = 0\), so angular momentum \(= 0\).

214. What is the maximum number of electrons in an atom that can have quantum numbers \(n = 3, l = 1\)?

Answer: 6

Explanation:
\(n=3, l=1\) corresponds to the \(3p\) subshell, which has 3 orbitals and can hold a maximum of 6 electrons.

215. According to Heisenberg's uncertainty principle, what is the product of uncertainty in position and momentum equal to or greater than?

Answer: h / (4\pi)

Explanation:
Heisenberg's Uncertainty Principle states that: $$\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$$

216. Which spectral series of the hydrogen atom lies in the visible region of the electromagnetic spectrum?

Answer: Balmer series

Explanation:
Transitions ending at \(n_1 = 2\) form the Balmer series, which falls in the visible light spectrum.

217. Calculate the wavenumber (\(\bar{\nu}\)) of the longest wavelength transition in the Lyman series of hydrogen atom. [\(R_H = 1.097 \times 10^7\,\text{m}^{-1}\)]

Answer: 8.2275 \times 10^6 m^{-1}

Explanation:
Longest wavelength in Lyman series occurs for \(n_1 = 1, n_2 = 2\): $$\bar{\nu} = R_H \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = \frac{3}{4} R_H = 0.75 \times 1.097 \times 10^7 = 8.2275 \times 10^6\,\text{m}^{-1}$$

218. What is the shape of a \(p\)-orbital?

Answer: Dumbbell

Explanation:
An \(s\)-orbital is spherical, a \(p\)-orbital is dumbbell-shaped, and a \(d\)-orbital is double-dumbbell-shaped.

219. How many angular nodes are present in a \(3d\) orbital?

Answer: 2

Explanation:
$$\text{Number of angular nodes} = l$$ For a \(d\) orbital, \(l = 2\), so it has 2 angular nodes.

220. What is the ground state electronic configuration of Copper (\(Z = 29\))?

Answer: [Ar] 3d^{10} 4s^1

Explanation:
Copper has a fully filled \(d\)-subshell for increased stability, promoting an electron from \(4s\) to \(3d\).

221. What is the velocity of an electron in the first orbit of hydrogen atom? [\(e = 1.6 \times 10^{-19}\,\text{C}, h = 6.626 \times 10^{-34}\,\text{J s}\)]

Answer: 2.18 \times 10^6 m/s

Explanation:
$$v_n = 2.18 \times 10^6 \times \frac{Z}{n}\,\text{m/s}$$ For \(n=1, Z=1\), \(v_1 = 2.18 \times 10^6\,\text{m/s}\).

222. Name the rule that forbids two electrons in an atom from having all four quantum numbers identical.

Answer: Pauli Exclusion Principle

Explanation:
Pauli's Exclusion Principle states that no two electrons in an atom can have the exact same set of four quantum numbers.

223. Calculate the radius of the second orbit of \(\text{Li}^{2+}\) ion.

Answer: 0.705 Å

Explanation:
$$r_n = 0.529 \times \frac{n^2}{Z}\,\text{Å}$$ For \(\text{Li}^{2+}\), \(Z = 3\), and for second orbit, \(n = 2\): $$r_2 = 0.529 \times \frac{4}{3} \approx 0.705\,\text{Å}$$

224. Which subshell has higher energy in a multi-electron atom: \(4s\) or \(3d\)?

Answer: 3d

Explanation:
By the \((n + l)\) rule:
For \(4s\): \(4 + 0 = 4\).
For \(3d\): \(3 + 2 = 5\).
Since \(5 > 4\), the \(3d\) subshell has higher energy.

225. What is the maximum number of unpaired electrons in \(\text{Fe}^{3+}\) (\(Z = 26\))?

Answer: 5

Explanation:
\(\text{Fe}\) ground state: \([\text{Ar}] 3d^6 4s^2\).
\(\text{Fe}^{3+}\) configuration: \([\text{Ar}] 3d^5\).
All 5 electrons in \(3d\) are unpaired.

226. What is the value of azimuth quantum number (\(l\)) for an \(f\)-subshell?

Answer: 3

Explanation:
\(l = 0\) for \(s\), \(l = 1\) for \(p\), \(l = 2\) for \(d\), and \(l = 3\) for \(f\).

227. Find the energy ratio of the first to the second orbit of hydrogen atom.

Answer: 4 : 1

Explanation:
$$E_n \propto \frac{1}{n^2} \implies \frac{E_1}{E_2} = \frac{2^2}{1^2} = \frac{4}{1}$$

228. What is the number of planar (angular) nodes in a \(4p\) orbital?

Answer: 1

Explanation:
Angular nodes \(= l\). For a \(p\) orbital, \(l = 1\), so it has 1 angular node.

229. Calculate the total number of nodes in a \(3p\) orbital.

Answer: 2

Explanation:
$$\text{Total nodes} = n - 1 = 3 - 1 = 2$$ (1 radial node + 1 angular node).

230. Which experiment proved the particle nature of electromagnetic radiation?

Answer: Photoelectric effect

Explanation:
The photoelectric effect (and the Compton effect) proved that light behaves as discrete packets of energy (photons), confirming its particle nature.

231. What is the kinetic energy of an photoelectron emitted when light of threshold frequency \(\nu_0\) strikes a metal surface?

Answer: h(\nu - \nu_0)

Explanation:
Einstein's photoelectric equation states: $$E = h\nu = h\nu_0 + \text{KE} \implies \text{KE} = h(\nu - \nu_0)$$

232. What is the ratio of species \(\text{He}^+\) and \(\text{H}\) for their ionization energies from ground state?

Answer: 4 : 1

Explanation:
$$\text{IE} \propto Z^2$$ $$\frac{\text{IE}(\text{He}^+)}{\text{IE}(\text{H})} = \frac{2^2}{1^2} = \frac{4}{1}$$

233. How many orbitals are present in an \(f\)-subshell?

Answer: 7

Explanation:
For \(f\)-subshell (\(l = 3\)), \(m_l\) ranges from \(-3\) to \(+3\), giving \(2l + 1 = 7\) orbitals.

234. Calculate the uncertainty in momentum of an electron if the uncertainty in its position is \(1\,\text{Å}\) (\(10^{-10}\,\text{m}\)).

Answer: 5.27 \times 10^{-25} kg m/s

Explanation:
$$\Delta p = \frac{h}{4\pi \Delta x} = \frac{6.626 \times 10^{-34}}{4 \times 3.1416 \times 10^{-10}} \approx 5.27 \times 10^{-25}\,\text{kg m/s}$$

235. State Hund's Rule of Maximum Multiplicity.

Answer: Electron pairing in degenerate orbitals cannot occur until each orbital is singly occupied.

Explanation:
Orbitals of equal energy (degenerate) are each occupied singly with parallel spins before any orbital is doubly occupied.

236. What is the value of spin angular momentum for a single electron?

Answer: \frac{\sqrt{3}}{2} \frac{h}{2\pi}

Explanation:
$$\text{Spin angular momentum} = \sqrt{s(s+1)} \frac{h}{2\pi}$$ For \(s = \frac{1}{2}\): \(\sqrt{\frac{1}{2}\left(\frac{3}{2}\right)}\frac{h}{2\pi} = \frac{\sqrt{3}}{2} \frac{h}{2\pi}\).

237. Which series of hydrogen spectrum lies in the infrared region and starts at \(n_1 = 3\)?

Answer: Paschen series

Explanation:
Transitions ending at \(n_1 = 3\) belong to the Paschen series, which falls in the infrared region.

238. Find the charge to mass ratio (\(e/m\)) of an electron.

Answer: 1.76 \times 10^{11} C/kg

Explanation:
$$\frac{e}{m} = \frac{1.602 \times 10^{-19}\,\text{C}}{9.109 \times 10^{-31}\,\text{kg}} \approx 1.76 \times 10^{11}\,\text{C/kg}$$

239. What is the maximum number of electrons in a shell with principal quantum number \(n\)?

Answer: 2n^2

Explanation:
The total capacity of a shell with quantum number \(n\) is given by Bohr-Bury scheme as \(2n^2\).

240. Calculate the wavelength of light emitted when an electron in hydrogen atom jumps from \(n = 2\) to \(n = 1\). [\(R_H = 1.097 \times 10^7\,\text{m}^{-1}\)]

Answer: 121.5 nm

Explanation:
$$\frac{1}{\lambda} = R_H \left(1 - \frac{1}{4}\right) = \frac{3}{4} R_H = 8.2275 \times 10^6\,\text{m}^{-1}$$ $$\lambda = \frac{1}{8.2275 \times 10^6} \approx 1.215 \times 10^{-7}\,\text{m} = 121.5\,\text{nm}$$

241. What is the designation for an orbital with \(n = 4\) and \(l = 2\)?

Answer: 4d

Explanation:
\(n=4\) denotes the 4th shell, and \(l=2\) corresponds to the \(d\) subshell, so the orbital designation is \(4d\).

242. State the Aufbau Principle.

Answer: Electrons fill subshells of lowest available energy levels before occupying higher levels.

Explanation:
In the ground state of an atom, atomic orbitals are filled in order of increasing energy levels according to the \((n+l)\) rule.

243. What is the number of unpaired electrons in a gaseous \(\text{Ni}^{2+}\) ion (\(Z = 28\))?

Answer: 2

Explanation:
\(\text{Ni}\) configuration: \([\text{Ar}] 3d^8 4s^2\).
\(\text{Ni}^{2+}\) configuration: \([\text{Ar}] 3d^8\).
Out of 8 electrons in five \(d\)-orbitals, 3 orbitals are doubly filled and 2 are singly filled (2 unpaired).

244. Determine the angular momentum of an electron in the 3rd orbit of hydrogen atom according to Bohr's theory.

Answer: 3h / (2\pi)

Explanation:
Bohr's quantization condition: \(L = \frac{n h}{2\pi}\).
For \(n = 3\), \(L = \frac{3h}{2\pi}\).

245. What is the name of the region around the nucleus where the probability of finding an electron is zero?

Answer: Nodal surface (or Node)

Explanation:
A node is a point or plane in an atom where the electron wave function \(\psi\) and \(\psi^2\) equal zero.

246. Calculate the de Broglie wavelength of a ball of mass \(0.1\,\text{kg}\) moving with a velocity of \(10\,\text{m/s}\).

Answer: 6.626 \times 10^{-34} m

Explanation:
$$\lambda = \frac{h}{m v} = \frac{6.626 \times 10^{-34}}{0.1 \times 10} = 6.626 \times 10^{-34}\,\text{m}$$

247. Which of the following orbitals does not exist: \(1p\), \(2s\), \(3f\), \(2p\)?

Answer: 1p and 3f

Explanation:
For \(p\)-subshell \(l=1\), so minimum \(n=2\) (\(1p\) cannot exist). For \(f\)-subshell \(l=3\), so minimum \(n=4\) (\(3f\) cannot exist).

248. Calculate the energy associated with the second Bohr orbit of \(\text{He}^+\).

Answer: -13.6 eV

Explanation:
$$E_n = -13.6 \times \frac{Z^2}{n^2}\,\text{eV}$$ For \(\text{He}^+\), \(Z = 2\); for 2nd orbit, \(n = 2\): $$E_2 = -13.6 \times \frac{2^2}{2^2} = -13.6\,\text{eV}$$

249. What are degenerate orbitals?

Answer: Orbitals belonging to the same subshell that have equal energy.

Explanation:
Orbitals of the same subshell (e.g., \(2p_x, 2p_y, 2p_z\)) sharing identical energy levels in the absence of an external magnetic field are termed degenerate.

250. Express Planck's constant in units of \(\text{eV s}\).

Answer: 4.135 \times 10^{-15} eV s

Explanation:
$$h = \frac{6.626 \times 10^{-34}\,\text{J s}}{1.602 \times 10^{-19}\,\text{J/eV}} \approx 4.135 \times 10^{-15}\,\text{eV s}$$