Question

A gas mixture contains 50% helium and 50% methane by volume. What is the percent by weight of methane in the mixture ?

• A. 0.1997
• B. 0.2005
• C. 0.5
• D. 0.8003

Answer 1

Answer: (D)

0.8003

Answer 2

PV = nRT
PV = $\frac{w}{M} RT$
V = $\frac{w}{M} \, \frac{RT}{P}$
Here R, T and P are constants
$\therefore$ $V_{He} \, \propto \frac{w(He)}{M(He)}$
and $V_{CH_4} \, \propto \frac{w(CH_4)}{M(CH_4)}$
But $V_{He}= V_{CH_4}$ (Given)
$\frac{w(He)}{M(He)} = \frac{w(CH_4)}{M(CH_4)}$
$\frac{w(He)}{4} = \frac{w(CH_4)}{16}$ \frac{w(He)}{M(CH_4)} = \frac{1}{4}$$w(CH_4)$= 4 w (He)$
Wt.% of $CH_4$ = $\frac{w(CH_4) \times 100}{w(CH_4) + w(He)}$
= $\frac{4 \times 100}{5}$ = 80%