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A vessel contains `0.1 mol e of He, 0.1` mole of `O_(2)` and `0.3` mole of `N_(2)`. The total pressure is `1` atomosphere. The pressure exerted by `O_(2)` isA. `380 "mm of Hg"`B. `456 "mm of Hg"`C. `304 "mm of Hg"`D. `152 "mm of Hg"` |
Answer» Correct Answer - D<br>Total moles `=0. 1(He)+0.1(O_(2))+0.3(N_(2))= 0.5 "moles"`<br> Pressure exerted by `O_(2)=` mole fraction of `O_(2)xx "total pressure"`<br> `(0.1)/(0.5)xx1=(1)/(5)xx760= 152 mm Hg` | |