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

Two liquids X and Y form an ideal solution. At 300 K, vapour pressure

1.	Two liquids X and Y form an ideal solution. At 300 K, vapour pressure of the solution containing 1 mole of X and 3 mole of Y is 550 mm Hg. At the same temperature, if 1 mole of Y is further added to this solution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure ( in mm Hg) of X and Y in their pure states will be, respectively :
Answer» <P>200and 300 300 and 400 400 and 600 500 and 300 Solution :According to Raoult.s law `p=p_(X)^(@)x_(X)+p_(Y)^(@)x_(Y)` Initially, `p=550` mm Hg, `x_(X)=1/4, x_(Y)=3/4` `550=p_(X)^(@)1/4+p_(Y)^(@)3/4` ...(i) On adding 1 MOLE of Y `p=560` mm Hg, `x_(X)=1/5, x_(Y)=4/5` `560=p_(X)^(@)1/5+p_(Y)^(@)4/5`...(ii) Solving eq. (i) and (ii), `p_(X)^(@)=400` mm Hg `p_(Y)^(@)=600` mm Hg.