Question

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

• A. $19.97\%$
• B. $20.05\%$
• C. $50\%$
• D. $80.03\%$

Answer 1

The correct option is D $80.03\%$

Solution contain $He+C{H}_4$

$\%$ by volume is same as $\%$ by moles.

Molecular wt. of $He=4g/mol$
Molecular wt. of $C{H}_4=16g/mol$

Assuming 1 mole of $He$ and 1 mole $C{H}_4$

$\%wt$ of $C{H}_4=\frac{wtofC{H}_4}{Totalwt}\times 100=\frac{16}{4+16}\times 100=80.0\%$