JEE Advanced 2021 — Paper 1 — Question 38

Molar mass of $\text{P}=40$ \underset{\left( 0.1\text{ }\!\!~\!\!\text{ mole} \right)}{\mathop{\text{P}}}\,\overset{75\text{ }\!\!%\!\!\text{ }\!\!~\!\!\text{ }}{\mathop{\to }}\,\underset{\left( 0.1\times \frac{3}{4} \right)}{\mathop{\text{Q}}}\, \underset{\left( 0.1\times \frac{3}{4} \right)}{\mathop{3{{Q}^{3}}}}\,\overset{40\text{ }\!\!%\!\!\text{ }\!\!~\!\!\text{ }}{\mathop{\to }}\,\underset{\left( \frac{1}{3}\times 0.1\times \frac{3}{4}\times 0.4 \right)}{\mathop{R}}\, So, moles of $\text{R}=0.01$ mole Molar mass of $\left( R \right)=162$ So, $x=0.01\times 162=1.62\text{ }\!\!~\!\!\text{ g}$