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 (Formula): Depression in freezing point is a colligative property, \[ \Delta T_f = K_f \times m = K_f \times \frac{w \times 1000}{M \times W} \] where \( w \) = mass of solute, \( W \) = mass of solvent in grams, \( M \) = molar mass of solute.
Step 2 (Rearrange for M): \[ M = \frac{K_f \times w \times 1000}{\Delta T_f \times W} \]
Step 3 (Given): \( K_f = 5.12 \), \( w = 1.00\ g \), \( W = 50\ g \), \( \Delta T_f = 0.40\ K \).
Step 4 (Substitution): \[ M = \frac{5.12 \times 1.00 \times 1000}{0.40 \times 50} \]
Step 5 (Arithmetic): Numerator \( = 5120 \); denominator \( = 20 \); \( M = 5120/20 = 256 \).
\[ \boxed{M = 256\ \text{g mol}^{-1}} \]