Question

The boiling point of a solution containing 1.5 g of dichlorobenzene in 100 g of benzene was higher by 0.268 K. Calculate the molar mass of dichlorobenzene. (\(K_b\) for benzene = 2.62 K molal\(^{-1}\))

Hint:

For boiling point elevation problems: \[ \Delta T_b = K_b \times m \] Always convert solvent mass into kilograms before calculating molality.

Answer 1

Step 1: Use the boiling point elevation formula: \[ \Delta T_b = K_b \times m \]
Step 2: Substitute the given values: \[ 0.268 = 2.62 \times m \] \[ m = \frac{0.268}{2.62} = 0.1023 \ \text{mol kg}^{-1} \]
Step 3: Mass of benzene (solvent) = 100 g = 0.1 kg Molality: \[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \] \[ \text{Moles of solute} = 0.1023 \times 0.1 = 0.01023 \]
Step 4: Calculate molar mass of dichlorobenzene: \[ \text{Molar mass} = \frac{\text{mass}}{\text{moles}} = \frac{1.5}{0.01023} \] \[ = 146.7 \approx 147 \ \text{g mol}^{-1} \]