Q. W g of a non-volatile electrolyte solid solute of molar mass M g mol−1 when dissolved in 100 mL water, decreases vapour pressure of water from 640 mm Hg to 600 mm Hg. If aqueous solution of the electrolyte boils at 375 K and Kb for water is 0.52 K kg mol−1, then the mole fraction of the electrolyte solute (x2) in the solution can be expressed as
(Given : density of water = 1 g/mL and boiling point of water = 373 K)
(A) \( \dfrac{16}{2.6} \times \dfrac{W}{M} \)
(B) \( \dfrac{1.3}{8} \times \dfrac{M}{W} \)
(C) \( \dfrac{2.6}{16} \times \dfrac{M}{W} \)
(D) \( \dfrac{1.3}{8} \times \dfrac{W}{M} \)
Explanation
Lowering in vapour pressure is given by Raoult’s law:
\[
\frac{\Delta P}{P^\circ} = x_2
\]
Here,
\[
\Delta P = 640 - 600 = 40 \text{ mm Hg}, \quad P^\circ = 640 \text{ mm Hg}
\]
\[
x_2 = \frac{40}{640} = \frac{1}{16}
\]
Elevation in boiling point:
\[
\Delta T_b = 375 - 373 = 2 \text{ K}
\]
Using boiling point elevation formula:
\[
\Delta T_b = i K_b m
\]
Mass of water = 100 mL = 100 g = 0.1 kg.
\[
m = \frac{W/M}{0.1} = \frac{10W}{M}
\]
Substitute values:
\[
2 = i \times 0.52 \times \frac{10W}{M}
\]
\[
i = \frac{2}{5.2} \times \frac{M}{W} = \frac{1}{2.6} \times \frac{M}{W}
\]
For electrolyte:
\[
x_2 = i \times \frac{n_2}{n_1}
\]
Moles of water:
\[
n_1 = \frac{100}{18} \approx 5.56
\]
Thus,
\[
x_2 = \frac{1}{2.6} \times \frac{W}{M} \times \frac{1}{5.56}
\]
\[
x_2 = \frac{1.3}{8} \times \frac{W}{M}
\]