JEE Chemistry Physical and Inorganic Chemistry MCQs Set 04

Answer: (a) isotropic and supercooled liquids

Explanation:
1. Amorphous solids have no long range order, so their properties are the same in all directions (isotropic).
2. They are regarded as supercooled liquids with a very high viscosity.

Teacher's Note:
1. Crystalline solids are anisotropic and have a sharp melting point.
2. Glass is the standard example of an amorphous solid.

Answer: (d) 5 : 2

Explanation:
1. Total charge on 8 sulphide ions is 16, so seven Fe atoms must carry +16.
2. Let x be Fe(II) and y be Fe(III): \(x+y=7\) and \(2x+3y=16\), which gives \(y=2\), \(x=5\).
3. Ratio \(= 5:2\).

Teacher's Note:
1. Balance the total positive charge against the total negative charge.
2. Solve the two equations for the atom counts.

Answer: (b) \(\dfrac{\sqrt3\,a}{2}\)

Explanation:
1. In CsCl (bcc type arrangement) the ions touch along the body diagonal.
2. Body diagonal \(=\sqrt3 a = 2(r^++r^-)\).
3. So \(r^++r^- = \dfrac{\sqrt3\,a}{2}\).

Teacher's Note:
1. Cs+ sits at the body centre and Cl- at the corners.
2. The coordination number in CsCl is 8:8.

Answer: (d) 5.77 \(\AA\)

Explanation:
1. In zinc blende the ions touch along the body diagonal one-quarter of the way: \(r^++r^- = \dfrac{\sqrt3}{4}a\).
2. \(a = \dfrac{4(0.7+1.8)}{\sqrt3} = \dfrac{10}{1.732} = 5.77\ \AA\).

Teacher's Note:
1. Zinc blende has fcc packing of anions with cations in half of the tetrahedral voids.
2. The coordination number is 4:4.

Answer: (b) Rate = k[A]\(^2\)

Explanation:
1. Doubling A gives four times the rate, so \(2^m = 4\) and the order in A is 2.
2. Doubling B has no effect, so the order in B is 0.
3. Rate = k[A]\(^2\).

Teacher's Note:
1. Compare the factor of change in rate with the factor of change in concentration.
2. A zero order species does not appear in the rate law.

Answer: (c) \((t_{100\%})_{\text{Reaction I}} : (t_{100\%})_{\text{Reaction II}} = 3 : 2\)

Explanation:
1. For zero order, the magnitude of the slope is the rate constant: \(k_I = \tan30^\circ = \dfrac1{\sqrt3}\) and \(k_{II} = \tan60^\circ = \sqrt3\).
2. Completion times: \(t_I = \dfrac{0.5}{1/\sqrt3} = \dfrac{\sqrt3}{2}\) min and \(t_{II} = \dfrac{1}{\sqrt3}\) min, so the ratio \(=\dfrac{\sqrt3/2}{1/\sqrt3} = \dfrac32 = 3:2\).
3. Checking the others: [A] = [C] at \(t=\dfrac{\sqrt3}{4}\) min where each is 0.25 M, so (d) is false; at that time [B] = 0.25 M and [D] = 0.75 M, so (b) is false, and when [B] = [A] = 0.25 M, [D] = 0.75 M is larger than [B], so (a) is false as well.

Teacher's Note:
1. In a concentration-time graph of a zero order reaction the slope is the rate constant.
2. Total time = initial concentration divided by k.

Answer: (b) \(t_{1/2}\propto a,\ t_{1/2}\propto a^0\)

Explanation:
1. For zero order, \(t_{1/2} = \dfrac{a}{2k}\), so it is proportional to a.
2. For first order, \(t_{1/2} = \dfrac{0.693}{k}\), which is independent of a (\(a^0\)).

Teacher's Note:
1. In general \(t_{1/2}\propto a^{1-n}\).
2. First order half life is constant, which is why radioactive decay has a fixed half life.

Answer: (a) 24%

Explanation:
1. In parallel first order reactions the products are formed in the ratio of their rate constants.
2. \(\%B = \dfrac{k_1}{k_1+k_2}\times100 = \dfrac{1.20}{1.20+3.8}\times100 = 24\%\).

Teacher's Note:
1. The branching ratio is independent of time.
2. Percent of C formed would be 76%.

Answer: (a) 53.6 kJ mol\(^{-1}\)

Explanation:
1. \(\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{T_2-T_1}{T_1T_2}\right)\).
2. \(\log2 = 0.301 = \dfrac{E_a}{2.303\times8.314}\times\dfrac{10}{300\times310}\).
3. \(E_a = 53598\) J mol\(^{-1}\) \(\approx 53.6\) kJ mol\(^{-1}\).

Teacher's Note:
1. Use the Arrhenius equation in two-temperature form.
2. Keep temperature in kelvin and R in joules.

Answer: (a) Exothermic

Explanation:
1. \(\Delta H = E_{a(f)} - E_{a(b)} = 30.5 - 45.4 = -14.9\) kJ/mol.
2. The negative value means heat is released, so the reaction is exothermic.

Teacher's Note:
1. The products lie lower in energy than the reactants when the backward barrier is higher.
2. For an endothermic reaction \(E_{a(f)} > E_{a(b)}\).

Answer: (b) \(r = k[A][B]\)

Explanation:
1. The slowest step is the rate determining step.
2. Step (i) involves one A and one B, so the rate is \(r = k[A][B]\).

Teacher's Note:
1. Intermediates such as C must not appear in the final rate law.
2. The overall equation does not decide the rate law.

Answer: (a) 20

Explanation:
1. The temperature rises by 20\(^\circ\)C, which is two intervals of 10\(^\circ\)C.
2. New rate \(= 5\times2^2 = 20\).

Teacher's Note:
1. The temperature coefficient of a reaction is the factor by which the rate changes for a 10 degree rise.
2. The factor is raised to the power of the number of 10 degree steps.

Answer: (a) 138.6 s

Explanation:
1. 20 g to 5 g is two half lives.
2. \(t = \dfrac{2.303}{k}\log\dfrac{20}{5} = \dfrac{2.303}{10^{-2}}\times0.602 = 138.6\) s.

Teacher's Note:
1. Each half life reduces the amount to half.
2. Here \(t_{1/2} = 69.3\) s so two half lives give 138.6 s.

Answer: (c) 271.6, 91.0 and 126.4 S cm\(^2\) mol\(^{-1}\)

Explanation:
1. Kohlrausch law: \(\Lambda_m^0 = \nu_+\lambda_+^0 + \nu_-\lambda_-^0\).
2. \(\ce{CaCl2}\): \(119.0 + 2\times76.3 = 271.6\).
3. \(\ce{CH3COONa}\): \(40.9+50.1 = 91.0\); NaCl: \(50.1+76.3 = 126.4\).

Teacher's Note:
1. Multiply each ionic conductivity by the number of such ions in the formula.
2. At infinite dilution the ions move independently.

Answer: (a) \(\ce{Ni^2+(aq) + Cu(s) -> Ni(s) + Cu^2+(aq)}\)

Explanation:
1. A reaction is spontaneous when \(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} > 0\).
2. For (a) the cathode is Ni (\(-0.25\) V) and the anode is Cu (0.34 V), so \(E^\circ_{cell} = -0.59\) V and it does not occur.
3. (b) 0.46 V, (c) 1.10 V and (d) 0.76 V are all positive.

Teacher's Note:
1. A metal displaces from solution any metal with a higher reduction potential.
2. Copper cannot displace nickel from its salt.

Answer: (d) 10

Explanation:
1. \(\Lambda_m = \dfrac{\kappa\times1000}{M}\) in S cm\(^2\) mol\(^{-1}\).
2. \(\Lambda_m = \dfrac{0.001\times1000}{0.1} = 10\) ohm\(^{-1}\) cm\(^2\) mol\(^{-1}\).

Teacher's Note:
1. The factor 1000 converts the concentration from mol/L to mol/cm\(^3\).
2. Molar conductance increases as the solution is diluted.

Answer: (b) On increasing the dilution, the specific conductance (K) decreases while the equivalent and molar conductances increase

Explanation:
1. On dilution the number of ions per unit volume falls, so specific conductance \(\kappa\) and the conductance of a given cell fall (so a and c are wrong).
2. Molar and equivalent conductances increase because of greater dissociation and less interionic attraction, so (d) is wrong and (b) is right.

Teacher's Note:
1. Conductivity (per unit volume) decreases on dilution but conductivity per mole increases.
2. For strong electrolytes the rise is small, for weak electrolytes it is large.

Answer: (a) \(\ce{SO2}\) is evolved

Explanation:
1. On discharge, Pb is oxidised to \(\ce{PbSO4}\) at the anode and \(\ce{PbO2}\) is reduced to \(\ce{PbSO4}\) at the cathode.
2. Overall: \(\ce{Pb + PbO2 + 2H2SO4 -> 2PbSO4 + 2H2O}\), so acid is consumed and water forms but no \(\ce{SO2}\) is evolved.

Teacher's Note:
1. The density of the electrolyte falls as the battery discharges.
2. Charging reverses the cell reaction.

Answer: (a) \(-644.6\)

Explanation:
1. \(E^\circ_{cell} = 1.23 + 0.44 = 1.67\) V.
2. For the doubled equation, \(n = 4\).
3. \(\Delta G^\circ = -nFE^\circ = -4\times96500\times1.67 \approx -644.6\) kJ.

Teacher's Note:
1. Multiplying the reaction by 2 doubles n and hence \(\Delta G^\circ\), but not \(E^\circ\).
2. A negative \(\Delta G^\circ\) shows rusting is spontaneous.

Answer: (a) 2.176 V

Explanation:
1. Efficiency \(= \dfrac{\Delta G^\circ}{\Delta H^\circ} = 0.70\), so \(\Delta G^\circ = -210\) kJ.
2. For one electron, \(E^\circ = \dfrac{210000}{96500} = 2.176\) V.

Teacher's Note:
1. Efficiency of a fuel or galvanic cell is the fraction of the enthalpy change converted to electrical work.
2. Here n = 1 because the reaction transfers one electron.

Answer: (c) Temperature

Explanation:
1. Vapour pressure is the pressure of vapour in equilibrium with the liquid at a given temperature.
2. It does not depend on the surface area or volume of the container, only on the nature of the liquid and the temperature.

Teacher's Note:
1. Vapour pressure increases rapidly with temperature.
2. A liquid boils when its vapour pressure equals the external pressure.

Answer: (b) Osmosis will take place in the opposite direction

Explanation:
1. The external pressure 4.0 atm is greater than the osmotic pressure 3.7 atm.
2. When the applied pressure exceeds the osmotic pressure, solvent flows from the solution to the pure solvent, which is reverse osmosis.

Teacher's Note:
1. Reverse osmosis is used for desalination of sea water.
2. Osmosis stops only when the applied pressure equals the osmotic pressure.

Answer: (d) All of these

Explanation:
1. Starting with 1 mole, after dissociation the total moles \(= (1-\alpha) + x\alpha + y\alpha\), so \(i = 1 + \alpha(x+y-1)\) (b is correct).
2. Rearranging gives \(\alpha = \dfrac{i-1}{x+y-1}\) (a), and multiplying numerator and denominator by \(-1\) gives \(\dfrac{1-i}{1-x-y}\) (c).
3. All three are equivalent.

Teacher's Note:
1. The factor i is the ratio of observed to calculated colligative property.
2. For association the van't Hoff factor falls below one.

Answer: (b) Benzene and methanol

Explanation:
1. In benzene and methanol, the benzene molecules break the hydrogen bonds of methanol, so the interactions in the mixture are weaker than in the pure liquids.
2. The vapour pressure is higher than that predicted by Raoult's law, so the deviation is positive.
3. The other pairs form stronger interactions and show negative deviation.

Teacher's Note:
1. Positive deviation goes with weaker A-B forces and a positive enthalpy of mixing.
2. Such solutions form minimum boiling azeotropes.

Answer: (d) All of the above

Explanation:
1. Mole fractions of all components of a solution add up to 1.
2. The ratio of mole fractions equals the ratio of moles because the total number of moles cancels.
3. The mole fraction of the solute is its moles divided by the total moles, so all three statements are correct.

Teacher's Note:
1. Mole fraction is dimensionless and independent of temperature.
2. The sum of all mole fractions is always unity.

Answer: (a) 200 ml and 800 ml

Explanation:
1. Let the volume of 2.5 N acid be V ml, so the other is \((1000-V)\) ml.
2. \(2.5V + 0.625(1000-V) = 1\times1000\), so \(1.875V = 375\) and \(V = 200\) ml.
3. The volumes are 200 ml and 800 ml.

Teacher's Note:
1. The number of equivalents is conserved on mixing: \(N_1V_1+N_2V_2 = N_3V_3\).
2. The answer lies closer to the weaker solution because 1 N is nearer to 0.625 N.

Answer: (a) 16.5 V

Explanation:
1. 5 g per 100 mL is 50 g per litre; molarity \(= \dfrac{50}{34} = 1.47\) M, so normality \(= 2.94\) N.
2. Volume strength \(= 5.6\times N = 5.6\times2.94 \approx 16.5\) V.

Teacher's Note:
1. Volume strength is the litres of oxygen at STP released by 1 litre of the solution.
2. Volume strength is 11.2 times the molarity.