Question

The freezing point of an aqueous solution of KCN containing 0.175molkg−1Kf was −0.7oCH2O=2kgmol−1K. On adding 0.095 mol of M(CN)2, the freezing point of the solution was –0.570C. Take the cryoscopic constant for water to be equal to 2 with the usual units. If the complex formation takes place according to the equation, M(CN)2+mKCN⇌Km[M(CN)m+2] what is the value of m (integral value) in the formula of the complex?

Answer 1

$K_f$ for $H_{2}O=2kgmo{l}^{-1}K$

$\Delta T_f$(KCN solution)=0.7

m' (molality of KCN solution)

$=0.175molk{g}^{-1}$

$△T_f=K_f\times m^{\prime }\times i$

$i=(1+x)$

$=\frac{\Delta T_f}{K_f\times m^{\prime }}=\frac{0.7}{2\times 0.175}$ =2.0

This gives, x=1, indicating 100% ionisation of KCN

$\Delta T_f$(of the complex)=0.57

$m^{\prime }\left(M\right(CN{)}_2)=0.095molk{g}^{-1}$

$K_m\left[M\right(CN{)}_{m+2}]\rightleftharpoons m{K}^{+}+[M(CN{)}_{m+2}{]}^{-}$

$\therefore i=1+(m-1)\alpha$

$=1+(m+1-1)\alpha$

$=1+m\alpha =1+m$

$i=(1+m)=\frac{△T_f}{K_f{m}^{\prime }}$

$=\frac{0.570}{2\times 0.095}=3$

$\therefore m=2$