# 100 g of liquid A

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 $A, n_{\mathrm{A}}=\frac{100}{140} \mathrm{~mol}$

= 0.714 mol

Number of moles of liquid $\mathrm{B}, n_{\mathrm{B}}=\frac{1000}{180} \mathrm{~mol}$

= 5.556 mol

Then, mole fraction of $\mathrm{A}, x_{\mathrm{A}}=\frac{n_{\mathrm{A}}}{n_{\mathrm{A}}+n_{\mathrm{B}}}$

$=\frac{0.714}{0.714+5.556}$

= 0.114

And, mole fraction of B, xB = 1 − 0.114

= 0.886

Vapour pressure of pure liquid $\mathrm{B}, p_{\mathrm{B}}^{0}=500$ torr

Therefore, vapour pressure of liquid B in the solution,

$p_{\mathrm{B}}=p_{\mathrm{B}}^{0} x_{\mathrm{B}}$

= 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,

$p_{\mathrm{A}}=p_{\mathrm{A}}^{0} x_{\mathrm{A}}$

$\Rightarrow p_{\mathrm{A}}^{0}=\frac{p_{\mathrm{A}}}{x_{\mathrm{A}}}$

$=\frac{32}{0.114}$

= 280.7 torr

Hence, the vapour pressure of pure liquid A is 280.7 torr.