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A gas mixture contains `50%` helium and `50%` methane by volume. What is the percent by weight of methane in the mixture.A. `80.03%`B. `20.05%`C. `50%`D. `75%`

Answer» Correct Answer - a
Solution contain `He+CH_(4)`
Their `mol. Wt.=4+16=20`
`% wt.` of `CH_(4)=(wt. of CH_(4))/("Total wt". )xx100=(16)/(20)xx10=80%`


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