1.

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

Answer»

0.1997
0.2005
0.5
`80.03%`

Solution :Molar and volume ratio will be same. i.e., 1:1
`therefore` Mass of 1 MOLE `CH_(4)` and He will be 16 and 4 g respectively.
Percentage by mass of `CH_(4)=("Mass of"CH_(4))/("Total mass")XX100`
`=(16)/(20)xx100=80%`


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