When 1 g each of compounds $A B$ and $A B_{2}$ are dissolved in 15 g of water separately, they incre

Question

When 1 g each of compounds $A B$ and $A B_{2}$ are dissolved in 15 g of water separately, they increased the boiling point of water by 2.7 K and 1.5 K respectively. The atomic mass of $A$ (in amu) is $\_\_\_\_$ $	imes 10^{-1}$. (Nearest integer)\ (Given: Molal boiling point elevation constant is $0.5 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$.)

Professor PC Thomas Classes · Answer

(25) : $\Delta T_{b}=K_{b} m$\ $m_{A B}=\frac{\Delta T_{b}}{K_{b}}=\frac{2.7}{0.5}=5.4$\ $m_{A B_{2}}=\frac{1.5}{0.5}=3$\ Moles of $A B 	imes \frac{1000}{15}=5.4$\ Moles of $A B=\frac{5.4 	imes 15}{1000}=0.081 \mathrm{~mol}$\ Likewise moles of $A B_{2}=\frac{3 	imes 15}{1000}=0.045$\ Mole $=\frac{	ext { Mass }}{	ext { Molar mass }}=\frac{1}{A_{A}+A_{B}}=0.081$\ $=\frac{1}{A_{A}+2 A_{B}}=0.045$\ $A_{A}+A_{B}=12.35 \mathrm{~g} / \mathrm{mol}$\ $A_{A}+2\left(A_{B}ight)=22.22 \mathrm{~g} / \mathrm{mol}$\ On solving, we get\ $A_{B}=9.87 \mathrm{~g} / \mathrm{mol}$\ $A_{A}=(12.35-9.87)=2.48 \mathrm{~g} / \mathrm{mol}$.\ $A_{A}=24.8 	imes 10^{-1} \mathrm{~g} / \mathrm{mol} \approx 25 	imes 10^{-1} \mathrm{~g} \mathrm{mol}^{-1}$\ or $25 	imes 10^{-1}$ amu