Question

Calculate the amount of $CaC{l}_2$(molar mass = 111g/mol) which must be added to 580g of water so that freezing point lower by 2K, assuming $CaC{l}_2$ is completely dissociated:

[$K_f$ for water = 1.86 K kg/mol].

Answer 1

As we know that,

$\Delta T_f=i\times \frac{K_f\times w_B\times 1000}{w_A\times M_B}.....\left(1\right)$

Given:-

$K_f$ for water $=1.86K.Kg/mol$

$M_B=111g/mol$

$w_A=580g$

$\Delta T_f=2K$

$w_B=?$

Dissociation of $Ca{Cl}_2$

$Ca{Cl}_2⟶{Ca}^{+2}+2{Cl}^{-}$

For the above reaction-

$i=3$

Substituting all these values in ${eq}^n\left(1\right)$, we have

$2=3\times \frac{1.86\times w_B\times 1000}{111\times 580}$

$\Rightarrow w_B=\frac{2\times 111\times 580}{1.86\times 3\times 1000}$

$\Rightarrow w_B=23.07g$

Hence the amount of $Ca{Cl}_2$ required is $23.07g$.