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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» 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%` |
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