Question

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

Answer 1

$CaC{l}_2\to C{a}^{2+}+2C{l}^{-}$

$\therefore i=3$

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

$2=3\times 1.86\times m$

$\therefore m=\frac{2}{3\times 1.86}=\frac{2}{5.58}=0.358$

$\therefore$ Amount of $CaC{l}_2$ to be added to
500 g of water $=\frac{0.358\times 500}{1000}\times 111$

=19.869 g.