Chemistry - Ideal Gas Question with Solution | TestHub

From $\mathrm{PV}=\mathrm{nRT}$, total moles $=0.6$.

Let ${\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_8$ be ' a ' mole therefore moles of ${\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_{12}=(0.6-\mathrm{a})$ mole

Mass of ' c ' in ' a ' mole of ${\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_8=12\mathrm{ax}\mathrm{gm}$.

Mass of ' c ' in ( $0.6-\mathrm{a}$ ) mole of ${\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_{12}=12(0.6-\mathrm{a})\mathrm{x}$. gm.

Therefore total mass of ' c ' in mixture $=12\mathrm{ax}+12(0.6-\mathrm{a})\mathrm{x}$

$=7.2\mathrm{x}=41.4\times \frac{87}{100}=5.00$