Explanation (Complete Step-by-Step Calculation)

Step 1: Use boiling point elevation formula

\[ \Delta T_b = K_b \times m \]

\[ 0.5 = 5.0 \times m \]

\[ m = \frac{0.5}{5} \]

\[ m = 0.1 \text{ mol kg}^{-1} \]

Step 2: Calculate moles of solute

Mass of solvent = 150 g = 0.150 kg

\[ m = \frac{n_{solute}}{kg_{solvent}} \]

\[ 0.1 = \frac{n}{0.150} \]

\[ n = 0.015 \text{ mol} \]

Step 3: Moles of solvent

Molar mass of solvent = 300 g mol⁻¹

\[ n_{solvent} = \frac{150}{300} \]

\[ = 0.5 \text{ mol} \]

Step 4: Relative lowering in vapour pressure

For dilute solution:

\[ \frac{\Delta P}{P^0} = X_{solute} \]

\[ X_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} \]

\[ = \frac{0.015}{0.015 + 0.5} \]

\[ = \frac{0.015}{0.515} \]

\[ \approx 0.029 \]

Express in form ______ × 10⁻2:

\[ 0.029 = 2.9 \times 10^{-2} \]

Nearest integer = 3

Final Answer = 3

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