Question

Calculate the amount of $CaC{l}_2$ (molar mass = 111 g $mo{l}^{-1}$) which must be added to 500 g of water to lower its freezing point by 2 K, assuming $CaC{l}_2$ is completely dissociated. ($K_f$ for water = 1.86 K kg $mo{l}^{-1}$)

Answer 1

Given:

Molar mass of $CaC{l}_2$ $\left(M_B\right)$ = 111 g/mol

Weight of water $\left(W_A\right)$ = 500 g

$K_f$ for water = 1.86 K kg/mol

$\delta T_f$ = 2 K

Formula: $\Delta T_f=\frac{K_f\times w_B\times 1000}{w_A\times M_B}$

Solution :

$\Delta T_f=\frac{K_f\times w_B\times 1000}{w_A\times M_B}$

$2=\frac{1.86\times w_B\times 1000}{500\times 111}$

$w_B=\frac{2\times 500\times 111}{1.86\times 1000}=59.68$

Amount of $CaC{l}_2$ required = 59.68 g