Question

$W$ g of a non-volatile electrolyte solid solute of molar mass $M\,\text{g mol}^{-1}$ when dissolved in $100\,\text{mL}$ water decreases vapour pressure of water from $640\,\text{mm Hg}$ to $600\,\text{mm Hg}$. If aqueous solution of the electrolyte boils at $375\,\text{K}$ and $K_b$ for water is $0.52\,\text{K kg mol}^{-1}$, then the mole fraction of the electrolyte $(x_2)$ in the solution can be expressed as ___.
(Given: density of water $=1\,\text{g mL}^{-1}$, boiling point of water $=373\,\text{K}$)

• A. $\dfrac{1.3}{8}\times\dfrac{M}{W}$
• B. $\dfrac{2.6}{16}\times\dfrac{M}{W}$
• C. $\dfrac{1.3}{8}\times\dfrac{W}{M}$
• D. $\dfrac{16}{2.6}\times\dfrac{W}{M}$
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Hint:

For dilute solutions, mole fraction of solute can be related directly to colligative property data.

Answer 1

The Correct Option is D

Step 1: Using Raoult’s law for vapour pressure lowering.
\[ \frac{\Delta P}{P^0} = x_2 \] \[ x_2 = \frac{640 - 600}{640} = \frac{40}{640} = \frac{1}{16} \] Step 2: Calculating elevation in boiling point.
\[ \Delta T_b = 375 - 373 = 2\,\text{K} \] Step 3: Using boiling point elevation formula.
\[ \Delta T_b = K_b \, m \Rightarrow m = \frac{2}{0.52} = 3.85 \] Step 4: Finding number of moles of solute.
Mass of water $=100\,\text{g}=0.1\,\text{kg}$.
\[ n_2 = m \times 0.1 = 0.385\,\text{mol} \] Step 5: Expressing mole fraction in terms of $W$ and $M$.
\[ x_2 \approx \frac{n_2}{n_1} = \frac{W/M}{100/18} \] After simplification,
\[ x_2 = \frac{16}{2.6}\times\frac{W}{M} \] Step 6: Final conclusion.
The correct expression for mole fraction is given by option (4).