100 g of liquid A (molar mass 140 g `"mol"^(-1)` ) was dissolved in 1000 g of liquid B (molar mass 180 g `"mol"^(-1)` ). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr.

Explanation

Number of moles of liquid A (solute) = `100/140 = 5/7 ` mole
Number of moles of liquid B (solvent) = `1000/180 = 50/9 `moles
` therefore ` Mole fraction of A in the solution `(x_A) = (5//7)/(5//7 + 50//9) = (5//7)/(395//63) = 5/7 xx 63/395 = 45/395 = 0.114`
` therefore `Mole fraction of B in the solution (`x_B`) = 1 – 0.114 = 0.886
Also, given `p_B^0` = 500 torr
Applying Raoult.s law and substituting the values, we have
`p_A =x_A p_A^0 = 0.114 xx p_A^0`
`p_B = x_B p_B^0 = 0.886 xx 500 = 443`torr
`p_("total") =p_A + p_B`
`475 = 0.114 p_A^0 + 443`
`p_A^0 = (475 - 443)/(0.114) = 280.7` torr
Vapour pressure of A in the solution `p_A = 0.114 xx 280.7` torr = 32 torr.