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 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 mole. If the volume of the container is 3L. What is the molecular weight of the unknown gas?

Explanation

Step I. Calculations of no. of moles of the unknown gas
According to available data , `P = 6 "bar" , V= 3` litres , `T = 27+273 = 300 K`
`R = 0.083 L "bar" K^(-1) mol^(-1)`
Total no. of moles of the two gases can be calculated with the help of ideal gas equation :
`PV = nRT` or `n = (PV)/(RT)`
`n = ((6 "bar") xx (3 L))/((0.083 L "bar" K^(-1) mol^(-1)) xx (300 K)) = 0.723 mol`
No. of moles of hydrogen `(n_(H_(2))) = 0.7` mol
No. of moles of gas `(n_(" gas")) = 0.723 - 0.7 = 0.023` mol.
Calculation of molar mass of the unknown gas.
According to Avagadro's law. volume of the gas diffused is proportional to the number of moles of the gas
Rate of diffusion of hydrogen `(r_(H_(2))) = (0.7)/(20) mol "min"^(-1)`
Rate of diffusion of gas `(r_("gas")) = (0.023)/(20) mol "min"^(-1)`
Molas mass of hydrogen `(M_(H_(2))) = 2`
Molar mass of the gas `(M_("gas")) = ?`
According to Graham's law of diffusion,
`(r_(H_(2)))/(r_("gas")) = sqrt((M_("gas"))/(M_(H_(2))))` or `(0.7)/(20) xx (20)/(0.023) = sqrt((M_("gas"))/(2 mol))`