Question

If $1\text{g}$ of solute dimerise upto $75\%$ in $100\text{g}$ of $H_{2}O$ and show depression in freezing point by ${0.093}^{\circ }C$. Find average molecular weight of solute (in $\text{g}$) after dimerisation.
(Given: The freezing point depression constant of water is $K_{f\left(H_{2}O\right)}=1.86{}^{\circ }C/m$)

Answer 1

$$\begin{array}{cccc}& 2A& \to & A_2\\ \text{Initial:}& 1& & 0\\ \text{Final:}& 1-0.75& & \frac{0.75}{2}\\ & =0.25& & 0.375\end{array}$$
$i=\frac{\text{Moles after association}}{\text{Mole before association}}=\frac{0.25+0.375}{1}=0.625\text{Now,}\Delta T_f=i\times K_f\times m\Rightarrow 0.093=0.625\times 1.86\times \frac{1}{M}\times \frac{1000}{100}\Rightarrow M=\frac{0.625\times 1.86\times 10}{0.093}=125\text{g}$
We know,
$i=\frac{\text{Normal molecular mass}}{\text{Abnormal molecular mass}}\Rightarrow 0.625=\frac{125}{\text{Abnormal molecular mass}}\Rightarrow \text{Abnormal molecular mass}=\frac{125}{0.625}=200\text{g}$