At 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 300K. Then the vapour pressure of A in its pure state is

Options

A. 300mm of Hg
B. 400mm of Hg
C. 500mm of Hg
D. 600mm of Hg

Answer

A

Explanation

`p_(t)=p_(A)x_(A)+p_(B)x_(B)`
`"In lst case, "500=(1)/(3)p_(A)+(2)/(3)p_(B)`
`"or "1500=p_(A)+2p_(B)` . . . (i)
in llnd case, (when one more mole of B isadded )
`2100=p_(A)+3p_(B)`
From Eqs. (i) and (ii)
`p_(B)=600mm" Hg and p_(A)=300mm Hg"`