Two liquids X and Y form an ideal solution at $$300$$ K, vapour pressure of the solution containing $$1$$ mol of X and $$3$$ mol of Y is $$550$$ mmHg. At the same temperature, if $$1$$ mol of Y is further added to this solution, vapour pressure of the solution increases by $$10$$ mmHg. Vapour pressure (in$$ $$mmHg) of X and Y in their pure states will be, respectively

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Infinity Learn · Accepted Answer

$$pT=pAoXA+pBoXB550=pAo$$×$$14+pBo$$×$$34Thus$$, $$pAo+3pBo=2200$$                        … (i) When $$1$$ mole of Y is further added to the solution, vapour pressure of a solution becomes $$560$$ mm Hg. Thus, $$560=pAo$$×$$15+pBo$$×$$45Thus$$, $$pAo+4pBo=2800$$                          … (ii)On$$ subtracting Eq. $$(ii) by Eq. (i) we get $$pBo=2800-2200$$⇒$$pBo=600Putting$$ the value of pBo in Eq. (i) $$pAo+3$$×$$600=2200pAo=2200-1800=400$$

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