Question

The freezing point decreases by 0.40 K, when 1.00 g non-electrolyte is dissolved in 50 g benzene. The value of \( K_f \) for benzene is 5.12 K kg mol\(^{-1}\). Calculate the molar mass of the solute.

Hint:

Use \( M=\dfrac{K_f\,w\,1000}{\Delta T_f\,W} \) with \( w=1 \) g, \( W=50 \) g.

Answer 1

Step 1: First find the molality from \(\Delta T_f=K_f\,m\), so \(m=\dfrac{\Delta T_f}{K_f}=\dfrac{0.40}{5.12}=0.078125\ mol\,kg^{-1}\).
Step 2: Molality is moles of solute per kg of solvent. Here solvent \(=50\ g=0.050\ kg\), so moles of solute \(=m\times 0.050=0.078125\times0.050=3.906\times10^{-3}\ mol\).
Step 3: Molar mass \(=\dfrac{\text{mass}}{\text{moles}}=\dfrac{1.00}{3.906\times10^{-3}}\).
Step 4: \(M=256\ g\,mol^{-1}\).
\[ \boxed{M = 256\ \text{g mol}^{-1}} \]