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

Answer

A

Explanation

(i) `1//3 P_(A)^(@) + 2//3 P_(B)^(@) = 500 mm`
`:. P_(A)^(@) + 2P_(B)^(@) = 1500 mm`....(1)
(ii) `1//4 P_(A)^(@) + 3//4 P_(B)^(@) = 525 mm`
`P_(A)^(@) + 3P_(B)^(@) = 2100mm` ....(2)
Subtracting (1) from (2)
`P_(B)^(@) = 2100 - 1500 = 600 mm " of " Hg`
`:. P_(A)^(@) = (P_(A)^(@) + 2P_(B)^(@)) - 2P_(B)^(@)`
`:. 1500 - 600 xx 2 = 300 mm " of " Hg`