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.
Number of moles of liquid A, nA = \(\frac{100}{140}\)mol
= 0.714 mol
Number of moles of liquid B, nB = \(\frac{1000}{180}\)mol
= 5.556 mol
Then, mole fraction of A, xA = \(\frac{n_A}{n_A + n_B}\)
= \(\frac{0.714}{0.714+5.556}\)
= 0.114
And, mole fraction of B, xB = 1 - 0.114
= 0.886
Vapour pressure of pure liquid B, pB0 = 500 torr
Therefore, vapour pressure of liquid B in the solution,
pB = pB0xB
= 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, pA = pA0xA
⇒ pA0 = \(\frac{p_A}{x_A}\)
= \(\frac{32}{0.114}\)
= 280.7 torr
Hence, the vapour pressure of pure liquid A is 280.7 torr.