Question

Solute ‘X’ dimerises in water to the extent of 80%.2.5g of ‘X’ in 100g of water increases the boiling point by 0.3∘C. The molar mass of ‘X’ is [K_b​=0.52Kkgmol⁻¹]

• A. 26
• B. 13
• C. 52
• D. 65

Answer 1

The Correct Option is A

The formula for boiling point elevation is given by: \[ \Delta T_b = K_b \cdot m \cdot i \] where:
\( \Delta T_b \) is the boiling point elevation,
\( K_b \) is the ebullioscopic constant,
\( m \) is the molality of the solution,
\( i \) is the van't Hoff factor (number of particles produced per formula unit).
Given that \( \Delta T_b = 0.3^\circ \text{C} \), \( K_b = 0.52 \, \text{K kg mol}^{-1} \), and the solute 'X' dimerizes in water to the extent of 80%, the van't Hoff factor \( i = 1 + 0.8 = 1.8 \). Now, molality \( m \) is given by: \[ m = \frac{\text{mol of solute}}{\text{mass of solvent in kg}} \] The mass of the solvent is 100g, or 0.1kg. The number of moles of 'X' is: \[ \text{moles of X} = \frac{\text{mass of X}}{\text{molar mass of X}} = \frac{2.5}{M} \] where \( M \) is the molar mass of 'X'. Substitute into the boiling point elevation formula: \[ 0.3 = 0.52 \cdot \left( \frac{2.5}{M \cdot 0.1} \right) \cdot 1.8 \] Simplifying: \[ 0.3 = \frac{0.936 \cdot 2.5}{M} \] \[ M = \frac{0.936 \cdot 2.5}{0.3} = 26 \]
Thus, the molar mass of 'X' is 26 g/mol.

The correct option is (A) : 26