$\mathrm{A}+ 2\mathrm{B} + 3\mathrm{C}\rightleftharpoons {\mathrm{AB}}_2{\mathrm{C}}_3$ . Reaction of $6.0 \mathrm{g}$of atoms of $\mathrm{B}$, and $0.036 \mathrm{mol}$ of $\mathrm{C}$ yields $4.8 \mathrm{g}$ of compound ${\mathrm{AB}}_2{\mathrm{C}}_{3 }$. If the atomic masses of $\mathrm{A}$ and $\mathrm{C}$ are $60$ and $80 \mathrm{amu}$, respectively, the atomic mass $\mathrm{B}$ of is $(\mathrm{Avogadro} \mathrm{constant}= 6 \mathrm{x} {10}^{23} {\mathrm{mol}}^{-1})$

The given data are:

$\underset{6.0 \mathrm{g} }{\mathrm{A}}+\underset{6.0 \mathrm{x} {10}^{23}\mathrm{atoms} }{2\mathrm{B}}+\underset{0.036 \mathrm{mol}}{3\mathrm{C}}\equiv \underset{4.8 \mathrm{g}}{{\mathrm{AB}}_2{\mathrm{C}}_3}$

$=\frac{6}{60}\mathrm{mol}\simeq 1\mathrm{mol}$

Maximum extent of reaction $0.1\mathrm{mol}0.5\mathrm{mol}0.012\mathrm{mol}$

The limiting reagent is the atom$\mathrm{C}$. The amount of ${\mathrm{AB}}_2{\mathrm{C}}_3$ formed will also be 0.012 mol. Mass of ${\mathrm{AB}}_2{\mathrm{C}}_3$ formed is $4.8 \mathrm{g}$. Hence

$\left(0.012\times 60+(0.024mol)M_R+036\times 80\right)g\mathit{=}\mathit{4}\mathit{.8}g$

This gives  $M_{\mathrm{B}}=50\mathrm{g}{\mathrm{mol}}^{-1}$ Thus, atomic mass of B is $50 \mathrm{amu}$.