Question

What do you understand by elevation in boiling point? Boiling point of a liquid is 350 K. On dissolving 2.0 g of a non-volatile solute in 100 g liquid, the boiling point of solution becomes 350.50 K. Calculate the molar mass of the solute. \( K_b \) for the liquid is 2.50 K kg mol\(^{-1}\).

Hint:

The elevation in boiling point is directly proportional to the molality and the ebullioscopic constant of the solvent.

Answer 1

Step 1: Formula for Elevation in Boiling Point.
The elevation in boiling point \( \Delta T_b \) is given by the formula: \[ \Delta T_b = K_b \times m \] where: - \( \Delta T_b \) is the change in boiling point, - \( K_b \) is the ebullioscopic constant of the solvent (2.50 K kg mol\(^{-1}\)), - \( m \) is the molality of the solution. Step 2: Calculate the Change in Boiling Point.
Given that the boiling point of the solution is 350.50 K and the boiling point of the pure liquid is 350 K: \[ \Delta T_b = 350.50 - 350 = 0.50 \, \text{K} \] Step 3: Calculate the Molality.
Rearranging the formula to solve for molality: \[ m = \frac{\Delta T_b}{K_b} = \frac{0.50}{2.50} = 0.20 \, \text{mol/kg} \] Step 4: Calculate the Moles of Solute.
Molality is also defined as: \[ m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \] The mass of the solvent is 100 g = 0.1 kg. Substituting the values: \[ \text{moles of solute} = m \times \text{mass of solvent} = 0.20 \times 0.1 = 0.02 \, \text{mol} \] Step 5: Calculate the Molar Mass of the Solute.
The molar mass \( M \) is given by: \[ M = \frac{\text{mass of solute (g)}}{\text{moles of solute}} = \frac{2.0}{0.02} = 100 \, \text{g/mol} \]