A 300 K the vapour pressure of an ideal solution containing 1 mole of liquid A and 2 moles of liquid B is 500 mm of Hg. The vapour pressure of the solution increases by 25 mm of Hg if one more mole of B is added to the above ideal solution at 300 K. Then vapour pressure of A in its pure state is :

A 300 K the vapour pressure of an ideal solution containing 1 mole of

1.	A 300 K the vapour pressure of an ideal solution containing 1 mole of liquid A and 2 moles of liquid B is 500 mm of Hg. The vapour pressure of the solution increases by 25 mm of Hg if one more mole of B is added to the above ideal solution at 300 K. Then vapour pressure of A in its pure state is :
Answer» 300 mm of Hg 40 mm of Hg 500mm of Hg 600 mm of Hg. Solution :ACCORDING to Raoult.s law, `p=p_(A)^(@) x_(A)+p_(B)^(@) x_(B)` `p=500` mm Hg, `x_(A)=1/3, x_(B)=2/3` `500=1/3 p_(A)^(@)+2/3 p_(B)^(@)` or `1500=p_(A)^(@)+2p_(B)^(@)` ...(i) On adding 1 mole of B, `p=525` mm Hg, `x_(A)=1/4, " " x_(B)=3/4` `525=1/4 p_(A)^(@)+3/4p_(B)^(@)` SUBTRACTING (i) from (ii) `:.600 = p_(B)^(@)` `:.1500=p_(A)^(@)+2xx600` `:.p_(A)^(@)=300` mm Hg