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 :

Options

A. 300 mm of Hg
B. 40 mm of Hg
C. 500 mm of Hg
D. 600 mm of Hg.

Answer

A

Explanation

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