Question

A solute A dimerizes in water. The boiling point of a $2$ molal solution of A is $100.52°\mathrm{C}$. The percentage association of A is (round off to the nearest integer) (use ${\mathrm{K}}_{\mathrm{b}}$ for water $=0.52\mathrm{K}\mathrm{kg}{\mathrm{mol}}^{-1}$; boiling point of water $=100°\mathrm{C}$) -

Answer 1

Step 1: Elevation in boiling point for the solution -

$∆{\mathrm{T}}_{\mathrm{b}}={\mathrm{T}}_{\mathrm{b}}-{\mathrm{T}}_{\mathrm{b}}^0$ where, ${\mathrm{T}}_{\mathrm{b}}$ is the boiling point of the solution and ${\mathrm{T}}_{\mathrm{b}}^0$ is the boiling point of water.

$∆{\mathrm{T}}_{\mathrm{b}}=$$100.52-100$

$∆{\mathrm{T}}_{\mathrm{b}}=0.52$

Step 2: Calculation of percentage association-

$∆{\mathrm{T}}_{\mathrm{b}}={\mathrm{iK}}_{\mathrm{b}}\mathrm{m}$

Where, ${\mathrm{K}}_{\mathrm{b}}$(molal elevation constant) $=0.52$ (given)

m is the molality of the solution$=2$

Putting the values we get, $0.52=\mathrm{i}\times 0.52\times 2$

$\mathrm{i}=\frac{0.52}{0.52\times 2}$

$\mathrm{i}=\left(1-\frac{\alpha }{2}\right)$

$\frac{1}{2}=\left(1-\frac{\alpha }{2}\right)$

$\alpha =1$

Which means 100 % association.

So, the percentage association of A is 100 %.