At `27^(@)C`, hydrogen is leaked through a tiny hole into a vessel for `20 min`. Another unknown gas at the same temperature and pressure as that of hydrogen is leaked through the same hole for `20 min`. After the effusion of the gases, the mixture exerts a pressure of `6 atm`. The hydrogen content of the mixture is `0.7 mol`. If the volume of the container is `3 L`, what is the molecular weight of the unknown gas?

Options

A. 1033
B. 1050
C. 900
D. 980

Answer

A

Explanation

Using gas equation, PV = nRT
Total no. of moles of gases in the mixture (n)
`=(PV)/(RT)=(6xx3)/(0.0821xx300)=0.7308` moles.
Thus no. of moles of unknown gas= 0. 7308 - 0.7
= 0.0308 moles
Now we know that
`(r_(1))/(r_(2))=("moles of hydrogen gas")/("moles of unknown gas")=(0.7)/(0.0308)`
Also we know that `(r_(1))/(r_(2))=sqrt((M_(2))/(M_(1)))`
`thereforeM_(2)=((r_(1))/(r_(2)))^(2)M_(1)` or `M_(2)=((0.7)/(0.0308))^(2)xx2=1033`