Question

A 4 : 1 molar mixture of `He` and `CH_(4)` is contained in vessel at 20 per pressure. Due to a hole in the vessel the gas mixture leakes out. What is the compostion of mixture effusing out initially.
A. `33.3% He, 66.7% CH_(4)`
B. `66.7% He, 33.3% CH_(4)`
C. `40% He, 60% CH_(4)`
D. `60% He, 40% CH_(4)`

Answer 1

Correct Answer - B
`(r_(He))/(r_(CH_(4))) = sqrt((M_(CH_(4)))/(M_(He))) xx (P_(He))/(P_(CH_(4)))`
`P prop` moles
`(r_(He))/(r_(CH_(4))) = sqrt((16)/(4)) xx (4)/(1) = 2 xx 4 = 8.1`
Molar composition effusing = 8:1
Weight composition effustiing `= 8 xx 4 : 1 xx 16 = 32.:16`
% of `He = (32)/(48) xx 100 = 66.7%`
% `CH_(4) = 100 = 66.7 = 33.3%`