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A gaseous mixture contains 336cm^(3) of ...

A gaseous mixture contains `336cm^(3)` of `H_(2)` and `224cm^(3)` of He at STP. The mixture shows a pressure of 2 atm when it is kept in a container at `27^(@)C`. Calculate the volume of the gas?

Text Solution

Verified by Experts

At STP, `336cm^(3)` of `H_(2)-=(336)/(22400)-=0.015` mol of `H_(2)` and
`224cm^(3)` of `He-=(224)/(22400)-=0.01` mol of He.
`V=(nRT)/(P)=(0.015+0.01)xx(0.0821xx300)/(2)=307.8cm^(3)`
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