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A gas mixture contains 50% helium and 50% methane by volume .What is the percentage by mass of methane in the mixture

$\begin{array}{1 1}(a)\;19.97\%&(b)\;20.05\%\\(c)\;50\%&(d)\;80.03\%\end{array}$

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Molar and volume ratios will be same (i.e) 1 : 1
$\therefore$ Mass of 1 mole $CH_4$ and $He$ will be 16 and 4g respectively.
Percentage by mass of $CH_4=\large\frac{Mass\;of\;CH_4}{Total\; mass}$$\times 100$
$\qquad\qquad\qquad\qquad\quad\;\;=\large\frac{16}{20}$$\times 100$
$\qquad\qquad\qquad\qquad\quad\;\;=80\%$
Hence (d) is the correct answer.
answered Oct 23, 2013 by sreemathi.v
 

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