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

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