Question
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.
Solution
Number of moles of liquid A, `n_A = 100/140 "mol"`
= 0.714 mol
Number of moles of liquid B, `n_B = 1000/180 "mol"`
= 5.556 mol
Then, mole fraction of A, `x_A = n_A/(n_A+n_B)`
`= 0.714/(0.714+5.556)`
= 0.114
And, mole fraction of B, xB = 1 − 0.114
= 0.886
Vapour pressure of pure liquid B, `p_B^0`= 500 torr
Therefore, vapour pressure of liquid B in the solution,
`p_B= p_B^0x_B`
= 500 × 0.886
= 443 torr
Total vapour pressure of the solution, ptotal = 475 torr
∴ Vapour pressure of liquid A in the solution,
pA = ptotal − pB
= 475 − 443
= 32 torr
Now,
`p_A = p_A^0x_A`
`=>p_A^0 = p_A/x_A`
`= 32/0.114`
= 280.7 torr
Hence, the vapour pressure of pure liquid A is 280.7 torr.
