Ratio of $\frac{∆{\mathrm{T}}_{\mathrm{b}}}{{\mathrm{K}}_{\mathrm{b}}}$ of 6% ${\mathrm{AB}}_2$ and 9% ${\mathrm{A}}_2\mathrm{B}$ $\left({\mathrm{AB}}_2\&{\mathrm{A}}_2\mathrm{Bboth} \mathrm{are} \mathrm{non}-\mathrm{electrolytes}\right)$ is 1molal in both cases. Hence atomic masses of A and B are respectively

$\frac{∆{\mathrm{T}}_{\mathrm{b}}}{{\mathrm{K}}_{\mathrm{b}}}=m=1$

$\mathrm{m}=1=\frac{6}{{\mathrm{m}}_{\mathrm{A}}}\times \frac{1000}{100}\Rightarrow \mathrm{m}=60$

$\mathrm{m}=1=\frac{9}{{\mathrm{m}}_{\mathrm{A}}}\times \frac{1000}{100}\Rightarrow {\mathrm{m}}_{\mathrm{A}}=90$

${\mathrm{AB}}_2\mathrm{mol}.\mathrm{wt}=60\Rightarrow \mathrm{A}+2\mathrm{B}=60\to \left(1\right)$

${\mathrm{A}}_2\mathrm{B} \mathrm{mol}.\mathrm{wt}\Rightarrow 2\mathrm{A}+\mathrm{B}=90\to \left(2\right)$

$2\times \left(1\right)=2\mathrm{A}+4\mathrm{B}=120$

2 = 2A + B = 90

3B = 30 ⇒ B = 10

A + B = 60 ⇒ A + 10 = 60

∴A = 40