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.

Answer 1

The correct answer is: 280.7 torr
Number of moles of liquid A, $n_A=\frac{100}{140}mol$
=0.714mol
Number of moles of liquid B, $n_B=\frac{1000}{180}mol$
=5.556mol
Then, mole fraction of A, $x_A=\frac{n_A}{n_A+n_B}$
$=\frac{0.714}{0.714+5.556}$
=0.114
And, mole fraction of B, $x_B = 1 - 0.114$
= 0.886
Vapour pressure of pure liquid B, $p^o_B = 500 torr$
Therefore, vapour pressure of liquid B in the solution,
$p_B=p^o_Bx_B$
$= 500 × 0.886$
= 443 torr
Total vapour pressure of the solution, ptotal = 475 torr
∴Vapour pressure of liquid A in the solution,
$p_A = p_{total} - p_B$
= 475 - 443
= 32 torr
Now,
$p_A=p^o_Ax_A$
$⇒p^o_A=\frac{p_A}{x_A}$
$=\frac{32}{0.114}$
= 280.7 torr
Hence, the vapour pressure of pure liquid A is 280.7 torr