Question

The freezing point of an aqueous solution of $KCN$ containing $0.1892$ mol/kg H${}_2$O was $-{0.704}^o$C. On adding $0.095$ mole of $Hg(CN{)}_2$, the freezing point of the solution was $-{0.53}^o$C. $Hg(CN{)}_2$ is the limiting reactant.

$Hg(CN{)}_2+mC{N}^{-}\to Hg(CN{)}_{m+2}^{m-}$

The value of m is?

• A. $2$
• B. $3$
• C. $4$
• D. $0$

Answer 1

The correct option is A $2$

The chemical reaction

$Hg(CN{)}_2+mC{N}^{-}\to Hg(CN{)}_{m+2}^{m^{-}}$

Initial moles $0.095$ $0.1892$ $0$

Final moles $0$ $0.1892-0.095m$ $0.095$

As $KCN$ dissociates completely,

$K_f=\frac{\Delta T_f}{i\times m}=\frac{0.704}{2\times 0.1892}=1.86$

Total number of particles after addition of $Hg(CN{)}_2$

$=0.1892+(0.1892-0.095m)+0.095$

$=0.4734-0.095m$

Now, $K_f=\frac{\Delta T_f}{i\times m}$

$\Rightarrow 1.86=\frac{0.53}{0.4734-0.095m}$

$\Rightarrow m\approx 2$

Therefore, the correct option is A.