Question

A very small amount of a non volatile solute(that does not dissociate) is dissolved in $56.8$ cc of benzene (density $0.889$ g/cc). At room temperature the vapour pressure of this solution is $98.88$ mm Hg while that of benzene is $100$ mm Hg. Find the molality of this solution freezing temperature of this solution is $0.73$ degree lower than that of benzene, hat is the value of molal freezing point depression constant of benzene.

Answer 1

$\frac{P^0-P}{P^0}=\frac{w}{m}\times \frac{M}{W}$

$\Delta T_f=K_f\times m$

put the given values

$\frac{1000-98.88}{98.88}=\frac{w\times 78\times 1000}{m\times W\times 1000}$

$\frac{w\times 100}{m\times W}=\frac{1.12\times 1000}{78\times 98.8}$

$\Delta T_f=K_f\times m$

$0.73=K_f\times 0.1452$

$K_f=5.027Kmolalit{y}^{-1}$ ($\frac{w}{m}\times \frac{1000}{W}=Molality$)