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 $5.12Kkgmo{l}^{-1}$. 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 $M_2=\frac{K_f\times w_2\times 1000}{\Delta T_f\times w_1}$
where $w_1=$ weight of solvent $=50g$
$w_2=$ weight of solute $=1.00g$
$K_f=$ molal depression constant $=5.12Kkgmo{l}^{-1}$
Substituting the values of various terms involved in the above equation, we get
$M_2=\frac{5.12\times 1.00\times 1000}{0.40\times 50}=256gmo{l}^{-1}$
Thus, molar mass of the solute $=256gmo{l}^{-1}.$