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

At 27^(@)C, hydrogen is leaked through a tiny hole into a vessel for 2

1.	At 27^(@)C, hydrogen is leaked through a tiny hole into a vessel for 20 minute. Another unknown gas at the same temperature and pressure as that of hydrogen leaked through the same hole for 20 minutes. After the effusion of the gases the mixture exerts a pressure of 6 atmosphere. The hydrogen content of the mixture is 0.7 mole. If the volume of the container is 3 litre, what is the molecular mass of the unknown gas ?
Answer» Solution :Let `P_(H_(2)) and P_(x)` be the partial pressures of hydrogen and unknown gas respectively and n be the number of moles of unknown gas. `P_(H_(2)) =(0.7)/(3)xx0.821xx300` `P_(x) =(n)/(3) xx0.0821xx300` Adding both `P_(H_(2))+P_(x) =6=(1)/(3) xx0.0821xx300(0.7+n)` `n=0.0308` mole Applying LAW of diffusion `(0.7//20)/(0.0308//20)=sqrt((M)/(2))` or, M=1033