Question

1 g of an non- electrolyte solute (molar mass g/mol) was dissolved in g of benzene. if the freezing point depression constant , kf of benzene is K kg/mol, the freezing point of benzene will be lowered by

1) 0.2K

2) 0.4K

3) 0.3)

4) 0.5K

Answer 1

Weight of solute = w_A = 1 g
Weight of solvent = w_B = g
Molar mass of solute = M_B = g/mol
​K_f = K kg/mol

$$\begin{gathered} {\mathrm{\Delta T}}_{\mathrm{f}}=\frac{{\mathrm{K}}_{\mathrm{f}}\times {\mathrm{w}}_{\mathrm{B}}\times 1000}{{\mathrm{w}}_{\mathrm{A}}\times {\mathrm{M}}_{\mathrm{B}}} \\ {\mathrm{\Delta T}}_{\mathrm{f}}=\frac{\mathrm{K}k\mathrm{g}/\mathrm{mol}\times 1\mathrm{g}\times 1000}{\mathrm{g}\times \mathrm{g}/\mathrm{mol}} \\ \Delta T_f=\frac{Kkg/mol\times g}{kg\times g/mol} \\ {\mathrm{\Delta T}}_{\mathrm{f}}=\mathrm{K} \end{gathered}$$

Depression in freezing point = K