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When `2 g` of a gas `A` is introduced into an evacuated flask kept at `25^(@)C`, the pressure is found to be `1 atm`. If `3 g` of another gas `B` is then heated in the same flask, the total pressure becomes `1.5 atm`. Assuming ideal gas behaviour, calculate the ratio of the molecular weights `M_(A)` and `M_(B)`.A. `1 :3`B. `3:1`C. `2:3`D. `3:2` |
Answer» Correct Answer - A Moles `prop` Pressure `(2)/(Mw_(A)) prop 1 "atm"` Pressure of `B = 1.5 - 1 = 0.5 "atm"` `(3)/(Mw_(B)) prop 0.5 "atm"` `(3)/(Mw_(B)) xx (Mw_(A))/(2) = (0.5)/(1)` `(MW_(A))/(Mw_(B)) = 0.5 xx (2)/(3) = (1)/(3)` `(Mw_(A) : Mw_(B) = 1:3` |
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