JEE Chemistry Physical Chemistry Mixed Topics MCQs Set 01

Answer: (d) 1.6

Explanation:
1. Total moles = 0.1 + 0.2 + 0.2 = 0.5.
2. Partial pressures: \(\ce{PCl5}\) = \(\tfrac{0.1}{0.5}\times 2 = 0.4\) atm; \(\ce{PCl3}\) = \(\ce{Cl2}\) = \(\tfrac{0.2}{0.5}\times 2 = 0.8\) atm.
3. \(K_p = \dfrac{0.8 \times 0.8}{0.4} = 1.6\).

Teacher's Note:
1. Convert mole fractions to partial pressures using the total pressure first.
2. \(K_p\) for this reaction has units of atm, here 1.6 atm.

Answer: (b) \(A(g) \rightleftharpoons \tfrac{n}{3}B(g) + \tfrac{2n}{3}C(g)\)

Explanation:
1. In this formula \(n\) is the total number of moles of gaseous products formed from one mole of \(A\).
2. Option (b): \(\tfrac{n}{3} + \tfrac{2n}{3} = n\).
3. Option (a) gives \(\tfrac{5n}{6}\), (c) gives \(\tfrac{3n}{4}\) and (d) gives \(\tfrac{n}{2} + 1\), none of which equals \(n\).
4. So only (b) fits the equation.

Teacher's Note:
1. For \(A \rightleftharpoons\) products, total product moles must equal n for this form of the degree-of-dissociation formula.
2. Add the coefficients to test each option quickly.

Answer: (c) 0.1 M \(\ce{CH3COONa}\)

Explanation:
1. \(\ce{NaCl}\) (strong acid and strong base) gives a neutral solution, pH about 7.
2. \(\ce{NH4Cl}\) (weak base, strong acid) is acidic.
3. \(\ce{CH3COONa}\) (weak acid, strong base) is basic because acetate hydrolyses to give \(\ce{OH-}\).
4. \(\ce{CH3COONH4}\) is almost neutral since \(K_a\) and \(K_b\) are nearly equal.
5. The basic salt has the highest pH.

Teacher's Note:
1. Classify each salt by the strength of the acid and base it comes from.
2. Salts of a weak acid with a strong base are always basic.

Answer: (d) Phenolphthalein (pH range 8.2 to 10)

Explanation:
1. At the equivalence point the solution contains the salt of a weak acid, which is basic (pH above 7).
2. The indicator must change colour in this basic range, so phenolphthalein (8.2 to 10) is suitable.

Teacher's Note:
1. Choose an indicator whose range includes the pH at the equivalence point.
2. Weak acid with strong base uses phenolphthalein; weak base with strong acid uses methyl orange.

Answer: (c) \(\ce{NH3}\)

Explanation:
1. By Graham's law, rate of diffusion is proportional to \(1/\sqrt{M}\).
2. Molar masses: \(\ce{O2}\) 32, \(\ce{CO2}\) 44, \(\ce{NH3}\) 17, \(\ce{N2}\) 28.
3. \(\ce{NH3}\) has the lowest molar mass, so it diffuses fastest.

Teacher's Note:
1. The lighter the gas, the faster it diffuses at the same temperature and pressure.
2. Ammonia is also lighter than all common air gases.

Answer: (a) Only B

Explanation:
1. For an ideal gas \(PV = nRT\) is constant at a fixed temperature, so the plot of \(PV\) against \(P\) is a horizontal straight line.
2. Curve B is the horizontal line starting at 1 on the \(PV\) axis.
3. A and E first dip then rise, and C and D rise steadily, so they show real gases.

Teacher's Note:
1. A horizontal line in a PV versus P plot means Boyle's law is obeyed exactly.
2. Real gases show a dip (attraction dominates) and then a rise (repulsion dominates).

Answer: (d) \(\dfrac{1}{76}\)

Explanation:
1. Pure water boils at 373 K, so its vapour pressure there is 760 mm.
2. Raoult's law: \(\dfrac{p^0 - p}{p^0} = x_{\text{solute}}\).
3. \(x_{\text{solute}} = \dfrac{760 - 750}{760} = \dfrac{1}{76}\).

Teacher's Note:
1. Relative lowering of vapour pressure equals the mole fraction of the non-volatile solute.
2. At the normal boiling point the vapour pressure of the pure liquid is 1 atm = 760 mm.

Answer: (a) \(\alpha = \dfrac{3(1 - i)}{2}\)

Explanation:
1. Start with 1 mole of A. After association, moles of A left \(= 1 - \alpha\) and moles of \(A_3\) formed \(= \dfrac{\alpha}{3}\).
2. Total moles \(= 1 - \alpha + \dfrac{\alpha}{3} = 1 - \dfrac{2\alpha}{3}\), which equals \(i\).
3. So \(\alpha = \dfrac{3(1 - i)}{2}\).

Teacher's Note:
1. Association lowers the number of particles, so i is less than 1.
2. That makes \(1 - i\) positive and \(\alpha\) positive.

Answer: (a) 110.8 atm

Explanation:
1. Moles of acid \(= 90/60 = 1.5\). Mass of solution \(= 90 + 110 = 200\) g.
2. Volume \(= 200/1.2 = 166.7\) mL \(= 0.1667\) L.
3. Molarity \(C = 1.5/0.1667 = 9\) M.
4. \(\pi = iCRT = 0.5 \times 9 \times 0.0821 \times 300 = 110.8\) atm.

Teacher's Note:
1. Acetic acid dimerises in benzene, which is why i is 0.5.
2. Use the density to convert the mass of solution into volume.

Answer: (b) Volume

Explanation:
1. Extensive properties depend on the amount of matter, such as volume, mass, energy and heat capacity.
2. Temperature, refractive index and viscosity do not depend on the amount, so they are intensive.

Teacher's Note:
1. Test: halve the system.
2. If the property halves, it is extensive; if it stays the same, it is intensive.

Answer: (c) \(-3RT\)

Explanation:
1. \(\Delta H = \Delta U + \Delta n_g RT\), so \(\Delta H - \Delta U = \Delta n_g RT\).
2. Gaseous products: 3 mol (water is liquid). Gaseous reactants: \(1 + 5 = 6\) mol.
3. \(\Delta n_g = 3 - 6 = -3\), so \(\Delta H - \Delta U = -3RT\).

Teacher's Note:
1. Count only gaseous species in \(\Delta n_g\).
2. Liquids and solids do not contribute.

Answer: (d) It increases with a decrease in pressure of the gas

Explanation:
1. Physisorption is exothermic, so it is favoured by low temperature, and (a), (b) and (c) are correct.
2. The extent of adsorption of a gas increases with an increase in pressure, not with a decrease.
3. So statement (d) is incorrect.

Teacher's Note:
1. Physisorption involves weak van der Waals forces and is reversible.
2. Chemisorption needs activation energy and is irreversible.

Answer: (b) Flocculation value

Explanation:
1. Flocculation value (coagulation value) is the minimum concentration, in millimoles per litre, of an electrolyte needed to coagulate a sol in a fixed time.
2. Gold number measures the protective power of a lyophilic colloid, not the coagulating power of an electrolyte.

Teacher's Note:
1. Smaller flocculation value means greater coagulating power.
2. Hardy-Schulze rule: higher charge on the effective ion gives more coagulating power.