Chemistry - Liquid Solution Question with Solution | TestHub

Given $\Delta T_f=0.3535$ K, $K_f=1.86$ K kg mol${}^{-1}$, $\alpha =0.9$, $M_B=74.5$ g/mol, $W_A=100$ g $=0.1$ kg.

For KCl, the number of ions $n=2$.

The van't Hoff factor $i$ is given by:

$i=1+(n-1)\alpha =1+(2-1)\times 0.9=1.9$

The depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=i\times K_f\times \frac{W_B}{M_B\times W_A}$

Substituting the given values:

$0.3535=1.9\times 1.86\times \frac{W_B}{74.5\times 0.1}$

$0.3535=3.534\times \frac{W_B}{7.45}$

$W_B=\frac{0.3535\times 7.45}{3.534}\approx 0.1\times 7.45=0.745$ g

Answer: $0.745$ g