Question

\(10\mathrm{g}\) of non-volatile solute is dissolved in \(180\mathrm{g}\) of \(\mathrm{H_2O}\) resulting in lowering of vapour pressure by \(0.5%\). Determine the boiling point of solution if \(K_b\) of water is \(0.52\mathrm{\ K\ kg\ mol^{-1}}\).

• A. \(100.01^\circ\mathrm{C}\)
• B. \(100.15^\circ\mathrm{C}\)
• C. \(100.23^\circ\mathrm{C}\)
• D. \(100.32^\circ\mathrm{C}\)

Hint:

Relative lowering of vapour pressure equals mole fraction of solute.

Answer 1

The Correct Option is B

Step 1: Formula / Definition}
\[ \frac{p^\circ - p}{p^\circ} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1} \]
Step 2: Calculation / Simplification}
\(0.005 = \frac{n_2}{180/18} = \frac{n_2}{10} \Rightarrow n_2 = 0.05\)
Molality \(m = \frac{0.05}{0.180} = 0.2778\)
\(\Delta T_b = K_b \cdot m = 0.52 \times 0.2778 = 0.1444 \approx 0.15\)
\(T_b = 100 + 0.15 = 100.15^\circ\mathrm{C}\)
Step 3: Final Answer
\[ 100.15^\circ\mathrm{C} \]