Mock Test 3: MCQs with Explanations

1. Which of the following statements is incorrect?

• (a) The conductance of one $cm^{3}$ (or 1 unit³) of a solution is called specific conductance.
• (b) Specific conductance increases while molar conductivity decreases on progressive dilution.
• (c) The limiting equivalent conductivity of weak electrolyte cannot be determined exactly by extrapolation of the plot of $\Lambda_{eq}$ against $\sqrt{c}$.
• (d) The conductance of metal is due to the movement of free electrons.

Correct Answer: (b)

Explanation: On dilution, the number of ions per unit volume decreases, leading to a decrease in specific conductance (conductivity). Conversely, molar conductivity increases with dilution due to the increase in the total volume of the solution.

2. In a cubic lattice each edge of the unit cell is 400 pm. Atomic weight of the element is 60 and its density is 6.25 g/c.c. The crystal lattice is:

• (a) face-centred
• (b) primitive
• (c) body-centred
• (d) end-centred.

Correct Answer: (a)

Explanation: Using the density formula $\rho = \frac{Z \cdot M}{N_A \cdot a^3}$, where $\rho = 6.25~g/cm^3$, $M = 60$, $N_A = 6 \times 10^{23}$, and $a = 400~pm = 4 \times 10^{-8}~cm$. Solving for Z: $6.25 = \frac{Z \times 60}{6 \times 10^{23} \times (4 \times 10^{-8})^3} \Rightarrow Z = 4$. A Z value of 4 corresponds to a face-centred cubic (fcc) lattice.

3. ZnS exists in two crystalline structures: (i) Zinc blende and (ii) Wurtzite. Mark the correct statement:

• (a) In zinc blende $S^{2-}$ ions are arranged in ccp while $Zn^{2+}$ occupy half of the tetrahedral sites.
• (b) In zinc blende $S^{2-}$ ions are arranged in hcp and $Zn^{2+}$ ions occupy all octahedral voids.
• (c) In Wurtzite structure $S^{2-}$ ions are arranged in ccp and $Zn^{2+}$ ions occupy half of tetrahedral voids.
• (d) In Wurtzite structures $S^{2-}$ ions are in hcp arrangement and $Zn^{2+}$ ions are occupying all octahedral voids.

Correct Answer: (a)

Explanation: In the zinc blende structure, sulphide ions form a cubic close-packed (ccp) lattice, and zinc ions occupy half of the available tetrahedral voids.

4. A solid AB has NaCl type structure. If the radius of cation $A^{+}$ is 170 pm, calculate the maximum possible radius of the anion $B^{-}$:

• (a) 210.3 pm
• (b) 397.4 pm
• (c) 410.6 pm
• (d) 347.9 pm

Correct Answer: (c)

Explanation: For an octahedral (NaCl type) structure, the radius ratio ($r_+/r_-$) must be between 0.414 and 0.732. To find the maximum radius of the anion, we use the minimum ratio: $\frac{170}{r_-} = 0.414 \Rightarrow r_- \approx 410.6$ pm.

5. During electrolysis, 1.05 g copper is deposited on electrode X. Mass of silver deposited on electrode Y would be:

• (a) 3.60 g
• (b) 1.80 g
• (c) 10.8 g
• (d) 1.05 g

Correct Answer: (a)

Explanation: According to Faraday's second law, equivalents of Ag deposited = equivalents of Cu deposited. $\frac{Mass_{Ag}}{108} = \frac{1.05}{63/2} \Rightarrow Mass_{Ag} = \frac{1.05 \times 2 \times 108}{63} = 3.60$ g.

6. Equivalent conductance of 1 M propanoic acid is 10 and at infinite dilution is 200. pH of the solution is:

• (a) 7
• (b) 3.3
• (c) 1.3
• (d) 6.8

Correct Answer: (c)

Explanation: Degree of dissociation $\alpha = \frac{\Lambda_{eq}}{\Lambda^0} = \frac{10}{200} = 0.05$. For 1 M acid, $[H^+] = C\alpha = 1 \times 0.05 = 0.05$ M. $pH = -\log(0.05) = 1.3$.

7. The equation for Langmuir adsorption isotherm under high pressure is:

• (a) $\frac{x}{m} = \frac{a}{b}$
• (b) $\frac{x}{m} = aP$
• (c) $\frac{x}{m} = \frac{1}{ap}$
• (d) $\frac{x}{m} = \frac{b}{a}$

Correct Answer: (a)

Explanation: In the Langmuir equation $\frac{x}{m} = \frac{ap}{1+bp}$, at high pressure, the term $bp$ is much larger than 1, so the equation simplifies to $\frac{x}{m} = \frac{a}{b}$ (constant).

8. In Langmuir's model of adsorption:

• (a) rate of dissociation does not depend on surface covered
• (b) adsorption at a single site may involve multiple molecules
• (c) mass of gas striking a given area is proportional to pressure
• (d) mass of gas striking a given area is proportional to temperature.

Correct Answer: (c)

Explanation: Langmuir's model assumes that the mass of the gas striking a unit area of the adsorbent surface per second is directly proportional to the pressure of the gas.

9. Equation describing change of $[N_{2}O_{5}]$ with time for its decomposition (first order):

• (a) $[N_{2}O_{5}]_{t}=[N_{2}O_{5}]_{0}+kt$
• (b) $[N_{2}O_{5}]_{0}=[N_{2}O_{5}]_{t}e^{kt}$
• (c) $log[N_{2}O_{5}]_{t}=log[N_{2}O_{5}]_{0}+kt$
• (d) $ln\frac{[N_{2}O_{5}]_{0}}{[N_{2}O_{5}]_{t}}=kt$

Correct Answer: (d)

Explanation: For a first-order reaction, the integrated rate law is $\ln([A]_0/[A]_t) = kt$, where $[A]_0$ is initial concentration and $[A]_t$ is concentration at time $t$.

10. The two-third life ($t_{2/3}$) of a first order reaction:

• (a) $\frac{2.303}{k} \log 3$
• (b) $\frac{2.303}{k} \log 2$
• (c) $\frac{2.303}{k} \log \frac{1}{3}$
• (d) $\frac{2.303}{k} \log \frac{2}{3}$

Correct Answer: (a)

Explanation: $t_{2/3}$ is the time when $x = \frac{2}{3}a$. Using the first-order equation: $t = \frac{2.303}{k} \log\frac{a}{a - (2/3)a} = \frac{2.303}{k} \log(3)$.

11. Heat released when 0.5 mol $HNO_3$ is added to 0.2 mol NaOH (standard heat = 57 kJ):

• (a) 11.4 kJ
• (b) 34.7 kJ
• (c) 23.5 kJ
• (d) 58.8 kJ

Correct Answer: (a)

Explanation: In neutralization, heat depends on the moles of $H_2O$ formed. Here, NaOH is the limiting reagent (0.2 mol). Heat = $0.2 \times 57 = 11.4$ kJ.

12. Unit cell with $\alpha=\beta=\gamma=90^{\circ}$ and $a=b\ne c$ is:

• (a) cubic
• (b) triclinic
• (c) hexagonal
• (d) tetragonal.

Correct Answer: (d)

Explanation: The described dimensions define a tetragonal crystal system.

13. Metal bcc lattice, edge 300 pm, molar mass 50. Density is:

• (a) 3.1 g/cm3
• (b) 6.2 g/cm3
• (c) 9.3 g/cm3
• (d) 12.4 g/cm3

Correct Answer: (b)

Explanation: For bcc, $Z=2$. $\rho = \frac{2 \times 50}{6.02 \times 10^{23} \times (3 \times 10^{-8})^3} \approx 6.15~g/cm^3 \approx 6.2~g/cm^3$.

14. Calculate $\Lambda_{m}^{\circ}[Ba(OH)_{2}]$ given NaOH, NaCl, $BaCl_{2}$ data:

• (a) $5.232\times10^{-2}$
• (b) $9.654\times10^{-2}$
• (c) $4.016\times10^{-2}$
• (d) $1.145\times10^{-1}$

Correct Answer: (a)

Explanation: $\Lambda^0 [Ba(OH)_2] = \Lambda^0 [BaCl_2] + 2\Lambda^0 [NaOH] - 2\Lambda^0 [NaCl]$. Substituting the values: $(2.8 + 2 \times 2.481 - 2 \times 1.265) \times 10^{-2} = 5.232 \times 10^{-2}~S~m^2~mol^{-1}$.

15. In adsorption graphs (A) and (B):

• (a) A is Freundlich, B is Langmuir
• (b) A is Langmuir, B is Freundlich
• (c) Both apply to gases on solids only
• (d) High pressure allows further adsorption.

Correct Answer: (b)

Explanation: Graph A shows saturation at high pressure (Langmuir), while B shows a continuous increase (Freundlich power law).

16. Adsorption theory speed increases because:

• (a) high concentration of reactants at active centres
• (b) $E_a$ becomes larger
• (c) heat increases speed
• (d) adsorption lowers $E_a$.

Correct Answer: (a)

Explanation: According to the adsorption theory, reactant molecules accumulate on the catalyst's surface, leading to a higher local concentration at the active sites, which increases the reaction rate.

17. In a Daniell cell:

• (a) Electrons flow Cu to Zn
• (b) Current flows Zn to Cu
• (c) Cations move toward Cu electrode
• (d) Cations move toward Zn electrode.

Correct Answer: (c)

Explanation: Zn is the anode and Cu is the cathode. During operation, cations from the salt bridge or solution move toward the cathode (Cu electrode) to maintain electrical neutrality as electrons arrive.

18. If forward $E_a=180$ and reverse $E_a=200$, $\Delta H$ is:

• (a) -20 kJ/mol
• (b) -300 kJ/mol
• (c) -120 kJ/mol
• (d) -280 kJ/mol

Correct Answer: (a)

Explanation: $\Delta H = E_{act(forward)} - E_{act(reverse)} = 180 - 200 = -20$ kJ/mol. Catalysts lower both equally, so $\Delta H$ remains unchanged.

19. Match substances with magnetic properties:

• (a) P-2, Q-4, R-1, S-3
• (b) P-4, Q-2, R-3, S-1
• (c) P-2, Q-4, R-3, S-1
• (d) P-4, Q-2, R-1, S-3

Correct Answer: (a)

Explanation: $CrO_2$ is ferromagnetic (2); $V_2O_5$ is diamagnetic (4); $V_2O_3$ is anti-ferromagnetic (1); TiO is paramagnetic (3).

20. While charging a lead storage battery:

• (a) $PbSO_4$ anode is reduced to Pb
• (b) $PbSO_4$ cathode is reduced to Pb
• (c) $PbSO_4$ cathode is oxidised to Pb
• (d) $PbSO_4$ anode is oxidised to $PbO_2$.

Correct Answer: (a)

Explanation: During charging, the external source reverses the cell reactions. $PbSO_4$ on the negative electrode (anode of the charger/cathode of the battery) is reduced to metallic Pb.

21. Products of electrolysis of aqueous $Na_{2}SO_{4}$ are:

• (a) Na and $SO_2$
• (b) $H_2$ and $SO_2$
• (c) Na and $O_2$
• (d) $H_2$ and $O_2$.

Correct Answer: (d)

Explanation: At the cathode, water is more easily reduced than $Na^+$ (producing $H_2$). At the anode, water is more easily oxidized than $SO_4^{2-}$ (producing $O_2$).

22. Least adsorbed gas on charcoal:

• (a) $CO_2$ (304 K)
• (b) $SO_2$ (630 K)
• (c) $CH_4$ (190 K)
• (d) $H_2$ (33 K).

Correct Answer: (d)

Explanation: Adsorption of a gas increases with its critical temperature. $H_2$ has the lowest critical temperature (33 K) and is thus the least adsorbed.

23. Rate equation for $P + 2Q \rightarrow 2R$:

• (a) $Rate = k[P]^2[Q]$
• (b) $Rate = k[P][Q]^2$
• (c) $Rate = k[P][Q]$
• (d) $Rate = k[P]^2[Q]^0$.

Correct Answer: (b)

Explanation: From Exp 1 and 3: doubling [P] doubles rate $\Rightarrow$ order is 1. From Exp 1 and 2: doubling [Q] quadruples rate $\Rightarrow$ order is 2. $Rate = k[P][Q]^2$.

24. In electrolysis of aqueous NaCl, anode reaction is:

• (a) $Na^+ + e^- \rightarrow Na$
• (b) $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$
• (c) $H^+ + e^- \rightarrow 1/2 H_2$
• (d) $Cl^- \rightarrow 1/2 Cl_2 + e^-$.

Correct Answer: (d)

Explanation: Due to overvoltage, the discharge of chloride ions to liberate chlorine gas is kinetically preferred over the discharge of oxygen from water.

25. Zero order plot (concentration vs time) has:

• (a) pos slope, zero intercept
• (b) neg slope, zero intercept
• (c) pos slope, non-zero intercept
• (d) neg slope, non-zero intercept.

Correct Answer: (d)

Explanation: $[A] = -kt + [A]_0$. The slope is $-k$ (negative) and the intercept is the initial concentration $[A]_0$ (non-zero).

26. Rubber in benzene is an example of:

• (a) multimolecular colloid
• (b) macromolecular colloid
• (c) associated colloid
• (d) lyophobic colloid.

Correct Answer: (b)

Explanation: High molecular weight polymers like rubber form macromolecular colloids when dissolved in suitable solvents.

27. Number of cation vacancies in NaCl doped with $10^{-3}$ mol $SrCl_2$:

• (a) $6.023\times10^{18}$
• (b) $6.023\times10^{20}$
• (c) $1.2\times10^{21}$
• (d) $3.011\times10^{20}$.

Correct Answer: (b)

Explanation: Each $Sr^{2+}$ ion replaces two $Na^+$ ions but occupies only one site, creating one vacancy. Total vacancies = $10^{-3} \times 6.023 \times 10^{23} = 6.023 \times 10^{20}$.

28. If $k = k_1 k_3 / k_2$, overall $E_a$ is:

• (a) 30 kJ
• (b) 15 kJ
• (c) 40 kJ
• (d) 60 kJ

Correct Answer: (c)

Explanation: From the Arrhenius equation, $E_a = E_{a1} + E_{a3} - E_{a2}$. Substituting values: $50 + 30 - 40 = 40$ kJ.

29. Time for 99.9% completion of first order:

• (a) 10 times $t_{1/2}$
• (b) 20 times $t_{1/2}$
• (c) 30 times $t_{1/2}$
• (d) 100 times $t_{1/2}$.

Correct Answer: (a)

Explanation: $t_{99.9} = \frac{2.303}{k} \log 1000 = \frac{6.909}{k}$. Since $t_{1/2} = \frac{0.693}{k}$, $t_{99.9} \approx 10 \times t_{1/2}$.

30. $k$ increases by 5% from 27 to 28°C. $E_a$ is:

• (a) 36.6 kJ/mol
• (b) 17.5 kJ/mol
• (c) 47.5 kJ/mol
• (d) 27.5 kJ/mol

Correct Answer: (a)

Explanation: Using $\log\frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left(\frac{T_2-T_1}{T_1 T_2}\right)$. $\log(1.05) = \frac{E_a}{2.303 \times 8.314} \left(\frac{1}{300 \times 301}\right) \Rightarrow E_a \approx 36.6$ kJ/mol.

31. Forward $E_a=249$, exothermic 30 kJ/mol. Reverse $E_a$ is:

• (a) 324 kJ/mol
• (b) 279 kJ/mol
• (c) 40 kJ/mol
• (d) 100 kJ/mol

Correct Answer: (b)

Explanation: $\Delta H = E_{a(f)} - E_{a(r)}$. $-30 = 249 - E_{a(r)} \Rightarrow E_{a(r)} = 279$ kJ/mol.

32. Density of CsBr (bcc lattice, edge 436.6 pm):

• (a) 4.25 g/cm3
• (b) 12.5 g/cm3
• (c) 0.425 g/cm3
• (d) 8.5 g/cm3.

Correct Answer: (d)

Explanation: [Note: Calculation usually yields ~4.25 g/cm³ for a body-centered cubic structure ($Z=1$ for CsBr). A $Z=2$ assumption yields 8.5].

33. 1L 1M $CuSO_4$ electrolysed with 2 F. Remaining molarity:

• (a) M/2
• (b) M/4
• (c) M/8
• (d) zero

Correct Answer: (d)

Explanation: Complete deposition of 1 mole of $Cu^{2+}$ requires 2 equivalents of electrons (2 F). Since the solution starts with 1 mole and exactly 2 F are passed, all copper is removed.

34. van't Hoff factor of very dilute $Ca(NO_3)_2$:

• (a) 1
• (b) 2
• (c) 3
• (d) 4

Correct Answer: (c)

Explanation: $Ca(NO_3)_2$ dissociates into one $Ca^{2+}$ and two $NO_3^-$ ions. In very dilute solutions, dissociation is nearly 100%, so $i = 3$.

35. Highest freezing point at 1 atm:

• (a) 0.1 M NaCl
• (b) 0.1 M sugar
• (c) 0.1 M $BaCl_2$
• (d) 0.1 M $FeCl_3$.

Correct Answer: (b)

Explanation: Freezing point is inversely proportional to the number of solute particles ($i$). Sugar ($i=1$) has the least depression and thus the highest freezing point.

36. Boiling point of 38.5 g solute (MW 154) in 250 g benzene ($K_b=2.61$, pure $bp=80.1$):

• (a) 85.32°C
• (b) 74.88°C
• (c) 77.49°C
• (d) 82.71°C.

Correct Answer: (d)

Explanation: $\Delta T_b = 2.61 \times \frac{38.5/154}{0.25} = 2.61^\circ C$. New $bp = 80.1 + 2.61 = 82.71^\circ C$.

37. Order of reaction $A_2 \rightleftharpoons 2A$ (fast), $A + B_2 \rightarrow AB + B$ (slow):

• (a) 2
• (b) 1
• (c) 1.5
• (d) 0

Correct Answer: (c)

Explanation: From equilibrium step, $[A] = K^{1/2} [A_2]^{1/2}$. From slow step, $Rate = k [A] [B_2] = k' [A_2]^{1/2} [B_2]^1$. Total order = $0.5 + 1 = 1.5$.

38. Inversion of cane sugar is:

• (a) second order
• (b) unimolecular
• (c) pseudo-unimolecular
• (d) none of these.

Correct Answer: (c)

Explanation: Although it involves two molecules, the large excess of water keeps its concentration effectively constant, resulting in first-order kinetics.

39. In autocatalysis:

• (a) reactant catalyses
• (b) product catalyses
• (c) solvent catalyses
• (d) heat catalyses.

Correct Answer: (b)

Explanation: Autocatalysis is a phenomenon where one of the reaction products acts as a catalyst for the reaction itself.

40. $A \rightarrow B$ second order. Doubling [A] increases rate by:

• (a) 2
• (b) 1/2
• (c) 4
• (d) 1/4.

Correct Answer: (c)

Explanation: $Rate = k[A]^2$. If $[A]$ is doubled, the new rate is $k(2[A])^2 = 4 \times k[A]^2$, which is a factor of 4 increase.

41. Lowest radius ratio for octahedral arrangement:

• (a) 0.155
• (b) 0.732
• (c) 0.414
• (d) 0.225.

Correct Answer: (c)

Explanation: For an octahedral void, the stable radius ratio range starts at 0.414.

42. 8:8 type packing is present in:

• (a) CsCl
• (b) KCl
• (c) NaCl
• (d) $MgF_2$.

Correct Answer: (a)

Explanation: CsCl exhibits a body-centered cubic structure with 8:8 coordination.

43. 0.108 g Ag deposited. Volume of oxygen evolved at STP:

• (a) 56 cm3
• (b) 550 cm3
• (c) 5.6 cm3
• (d) 11.2 cm3.

Correct Answer: (c)

Explanation: 0.108 g Ag = 0.001 equivalents. 0.001 equivalents of $O_2 = 0.001/4 = 0.00025$ moles. Vol = $0.00025 \times 22400~mL = 5.6~cm^3$.

44. Cell $Cd | Cd^{2+} || Pb^{2+} | Pb$ ($E^0_{Cd} = -0.403$, $E^0_{Pb} = -0.126$). $\Delta G$ is:

• (a) -53.46 kJ
• (b) 102.1 kJ
• (c) -102.1 kJ
• (d) 53.46 kJ.

Correct Answer: (a)

Explanation: $E_{cell} = -0.126 - (-0.403) = 0.277$ V. $\Delta G = -nFE_{cell} = -2 \times 96500 \times 0.277 \approx -53.46$ kJ.

45. Incorrect statement regarding physisorption:

• (a) van der Waals forces
• (b) Liquefiable gases adsorbed readily
• (c) High pressure yields multimolecular layer
• (d) Enthalpy is low and positive.

Correct Answer: (d)

Explanation: Adsorption is a spontaneous process, so $\Delta G$ and $\Delta S$ are negative, requiring the enthalpy change ($\Delta H$) to be negative (exothermic).