Question

1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is $5.12 \,K \,kg\, mol^{-1}$. The molar mass of the solute is

• A. 256 kg/mol
• B. 256 g mol
• C. 256 g/mol
• D. 256 mg/mol

Answer 1

Answer: (C)

We have $M_{2} = \frac{K_{f} \times w_{2}\times1000}{\Delta T_{f} \times w_{1}}$
where $w_{1} =$ weight of solvent $= 50\, g$
$w_{2} =$ weight of solute $= 1.00 \,g$
$K_{f } =$ molal depression constant $= 5.12\, K\, kg \,mol^{-1}$
Substituting the values of various terms involved in the above equation, we get
$M_{2} = \frac{5.12\times1.00\times1000}{0.40\times50} = 256 \,g \,mol^{-1}$
Thus, molar mass of the solute $= 256\, g \,mol^{-1}.$