From two identical holes, nitrogen and an unknown gas are leaked into a common vessel of `3 L` capacity for `10 min`, at `27^(@)C`. The resulting pressure is `4.18` bar and the mixture contains `0.4 mol` of nitrogen. What is the molar mass of the unknown gas?

Explanation

Step I. Calculation of number of moles of the unknown gas
For an ideal gas : `PV = n RT` or `n = (PV)/(RT)`
Volume of the container `(V) = 3 L = 3 dm^(3)`
Pressure of the gases present `(P) = 4.18 "bar"`
Temperature of the container ` (T) = 27+ 273 = 300 K`
Gas constant `(R) = 0.083 "bar" dm^(3) K^(-1) mol^(-1)`
`:.` Total no. of moles of the two gases `(n) = ((4.18 "bar") xx (3 dm^(3)))/((0.083 "bar" dm^(3) K^(-1) mol^(-1)) xx (300 K)) = 0.5036` mol
No. of moles of nitrogen `= 0.4` mol
No. of moles of unknown gas `= 0.5036 - 0.4 = 0.1036` mol
Step II. Calculation of molar mass of the unknown gas
Rate of effusion of nitrogen `(r_(N_(2))) = (0.4)/(10) mol "min"^(-1)`
Rate of effusion of the gas `(r_(g)) = (0.1036)/(10) mol "min"^(-1)`
According to Graham's Law, `(r_(N_(2)))/(r_(g)) = sqrt((M_(g))/(M_(N_(2)))) = sqrt((M_(g))/((28g mol^(-1))))`
or `(0.4)/((0.1036)) = sqrt((M_(g))/((28g mol^(-1))))`
squaring both sides, `M_(g) = (0.4 xx 0.4 xx (28 g mol^(-1)))/(0.1036 xx 0.1036) = 417.4 g mol^(-1)`.