Question

A substance x $\text{molecular weight}=94$ associates in water to form dimer. A solution of $1.25\text{g of}$ x in $50\text{g}$ of water lower the freezing point by ${0.3}^{\circ }\text{C}$, the degree of association of x is $(K_f=1.86)$

Answer 1

$\Delta T_f=i{K}_{f}m$
$0.3=i\times \frac{1.86\times 1.25\times 1000}{94\times 50}$
$\Rightarrow i=\frac{0.3\times 94\times 50}{1.86\times 1.25\times 1000}$
$\Rightarrow i=0.6$
$i=1+(\frac{1}{n}-1)\alpha$
$\Rightarrow 0.6=1+(\frac{1}{n}-1)\alpha$
$\Rightarrow 0.6=1+(\frac{1}{2}-1)\alpha$
$\Rightarrow 0.6=1+(-\frac{1}{2})\alpha$
$\Rightarrow 0.6=1-\frac{1}{2}\alpha$
$\Rightarrow \frac{1}{2}\alpha =0.4$
$\Rightarrow \alpha =0.8$