In a mixture of `O_(2)` and an unknown gas, the percentage of the unknown gas is 20% by mass. At a given temperature and pressure, time required to effuse V mL of the gas mixture through an aperrture is 234.1 s. under the same conditions, time required to effuse the same volume of pure `O_(2)` gas is 223.1 s. what is the molar mass of unknown gas?

Explanation

Average molecular mass of the gas-mixture `=(20xxM+80xx32)/(10)`, where M=molar mass of unknown gas.
According to the problem, the rate of effusion of V mL of unknown gas mixed with oxygen, `r_(1)=(V)/(234.1)mL*s^(-1)` and the rate of effusion of V mL of pure oxygen, `r_(2)=(V)/(223.1)mL*s^(-1)`
We know, `(r_(1))/(r_(2))=sqrt((M_(2))/(M_(1))) and (r_(1))/(r_(2))=sqrt((V)/(234.1))/((V)/(223.1))=(223.1)/(234.1)`
Now, `(M_(2))/(M_(1))=(32)/((20M+2560)/(100))=(160)/(M+128)`
`therefore (223.1)/(234.1)=sqrt((M_(2))/(M_(1)))=sqrt((160)/(M+128)) or, 0.9082=(160)/(M+128)`
`therefore M=48.17`
`therefore`Molar mass of unknown gas`=48.17g*mol^(-1)`