Question

$W \text{ gm}$ of non-volatile electrolyte solute is added in $100 \text{ ml}$ pure water ($\text{P}^\circ = 640 \text{ mm Hg}$) showing vapour pressure of solution $600 \text{ mm Hg}$. This solution have $\text{b}.\text{p}$. of $375 \text{ K}$. Given $K_b$ of $\text{H}_2\text{O} = 0.52 \frac{K \cdot \text{kg}}{\text{mol}}$. Molar mass of solute $= \text{M}$. Select the correct option about mole fraction of solute ($\text{X}_{\text{solute}}$).

• A. $\frac{1}{8} \frac{W}{M}$
• B. $\frac{2}{8} \frac{W}{M}$
• C. $\frac{2.6}{16} \frac{M}{W}$
• D. $\frac{1.3}{8} \frac{W}{M}$

Hint:

When combining data from different colligative properties, first derive the value of $W/M$ from the most reliable equation (here $\Delta T_b$), and then check which option provides the correct $X_{\text{solute}}$ derived from $\text{RLVP}$.

Answer 1

The Correct Option is D

Using Relative Lowering of Vapour Pressure ($\text{RLVP}$):
$X_{\text{solute}} = \frac{P^\circ - P_S}{P^\circ} = \frac{640 - 600}{640} = \frac{40}{640} = \frac{1}{16}$.
Using Boiling Point Elevation ($\Delta T_b$): $\Delta T_b = 375 \text{ K} - 373 \text{ K} = 2 \text{ K}$.
$\Delta T_b = i K_b m$. Assuming $i=1$. $m = \frac{W/M}{0.1 \text{ kg}}$.
$2 = 1 \times 0.52 \times \frac{W/M}{0.1}$.
$\frac{W}{M} = \frac{2 \times 0.1}{0.52} = \frac{0.2}{0.52} = \frac{1}{2.6}$.
We must find the option that equals $X_{\text{solute}} = 1/16$. We check Option (4):
Option (4) $=\frac{1.3}{8} \frac{W}{M}$. Substitute $W/M = 1/2.6$:
$\frac{1.3}{8} \times \frac{1}{2.6} = \frac{1.3}{20.8}$. Since $20.8 = 1.3 \times 16$:
$\frac{1.3}{20.8} = \frac{1.3}{1.3 \times 16} = \frac{1}{16}$.
Since $X_{\text{solute}}$ calculated from $\text{RLVP}$ is $1/16$, Option (4) is the correct symbolic representation.