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. 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 ` mmHg `x_(A) = (1)/(3) , x_(B) = (2)/(3)`
` 500 = (1)/(3) p_(A)^(@) + 2p_(B)^(@)`
or `" " 1500 = p_(A)^(@) + 2 p_(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)/(4) p_(B)^(@)` .
or `2100 = p_(A)^(@) + 3p_(B)^(@)" "....(ii)`
Subtractiong (i) form (ii) .
`" "600=p_(B)^(@)`
`therefore" "1500 = p_(A)^(@) + 2 xx600`
`therefore " "p_(A)^(@) = 300 mm Hg`