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JEE Chemistry Solved Question Paper 2006

Multiple Choice Questions

The relative lowering of vapour pressure of a dilute aqueous solution containing nonvolatile solute is 0.0125. The molality of the solution is about

0.70
0.50
0.90
0.80

If the elevation in boiling point of a solution of 10g of solute (mol. wt. = 100) in 100g of water is ΔT_b, the ebullioscopic constant of water is

100T_b
ΔT_b
10T_b
10

3.
An alloy of Pb-Ag weighing 1.08g was dissolved in dilute HNO₃ and the volume made to 100 mL. A silver electrode was dipped in the solution and the emf of the cell set up Pt(s), H₂(g)|H⁺(1M) || Ag⁺(aq)|Ag(s) was 0.62V. If E°_cell= 0. 80V what is the percentage of Ag in the alloy? [At 25°C, RT IF= 0.06].

50

1

10

25

Pt(s), H₂(g)|H⁺(1M) || Ag⁺(aq)|Ag(s)

EMF of cell = 0.62V, E°_cell= 0. 80V

$H_{2} \to 2 H^{+} + 2 e^{-} (at anode) \frac{2 Ag + 2 e^{-} \to 2 Ag (at cathode)}{H_{2} + 2 {Ag}^{+} \to 2 Ag + 2 H^{+}} E_{cell} = E ° - \frac{2.303 RT}{2 F} \log \frac{[H]^{+}}{[{Ag}^{+}]^{2} [H_{2}]} E_{cell} = E ° - \frac{2.303 RT}{2 F} \log \frac{1}{[{Ag}^{+}]^{2}} 0.62 = 0.80 + \frac{2.303 \times 0.06}{2} \log \frac{1}{[{Ag}^{+}]^{2}} 0.62 = 0.80 + \frac{2 \times 2.303 \times 0.06}{2} \log [Ag]^{+} - 0.80 = 0.1382 \log [Ag]^{+} [Ag]^{+} = 0.05 M ∴ Moles of {Ag}^{+} in 100 mL = 0.05 \times \frac{100}{1000} wt . of {Ag}^{+} in 100 mL = 0.05 \times \frac{100}{1000} \times 108 % of Ag in 1.08 g alloy = \frac{0.05 \times 100 \times 108}{1000 \times 1.08} \times 100 = 50 %$

The standard oxidation potentials of Zn, Cu, Ag and Ni electrodes are +0.76, 0.34, - 0.80 and +0.25 V respectively. Which of the following reaction will provide maximum voltage?