Two liquids X and Y on mixing form an ideal solution. The vapour pressure of the solution containing 3 mol of X and 1 mol of Y is 550 mm of Hg. But when 4 mol of X and 1 mol of Y are mixed, the vapour pressure of the solution, thus, formed is 560 mm of Hg. What will be the vapour pressure of the pure X and pure Y at this temperature?

Explanation

Let the vapour pressure of X be `p_1^@` and of Y be `p_2^@` and `X_1 and X_2` be their mole fractions.
Then according to Raoult.s law, the total pressure, P is
`P= p_1^@ * X_1 +p_2^@ * X_2`
In the first solution,
`X_1 =3/(3+1) =0.75, X_2 = 1/(3+2) = 0.25`
`:. p_1^@ xx 0.75 + p_2^@ xx 0.25 = 550 mm` .....(i)
In the second solution, `x_1= 4/(4+1) = 0.80, X_2 =1/(4+1) = 0.20`
`p_1^@ xx 0.80 + p_2^@ xx0.20 = 560 mm` ....(ii)
Multiplying eq. (i) by 4 and eq. (ii) by 5, we get
`3p_1^@ + p_2^@ = 2200 mm` ...(iii)
`4p_1^@ + p_2^@ = 2800 mm` ...(iv)
Subtracting eq. (iii) from eq. (iv)
`p_1^@ = 600 mm` of Hg
Substituting in eq. (iii) we get
`3 xx 600+ p_2^@ = 2200 or p_2^@ = 2200 - 1800 = 400 mm` Hg
Vapour pressure of pure component X = 600 mm Hg
Vapour pressure of pure component Y = 400 mm Hg