Question

A mixture contains 1 mole of volatile liquid A (PO A=100 mm Hg) and 3 moles of volatile liquid B (PO B=80 mm Hg). If the solution behaves ideally, the total vapour pressure of distillate is ???

Answer 1

Since the mixture behaves ideally we use Raoult's law, which states that the partial vapor pressure of the pure component is the initial vapor pressure of the pure component $(P⁰)$ at that temperature multiplied by the mole fraction in the mixture.

First we find the mole fractions for each liquid. The total number of moles is 4, so $n_A$, moles of liquid A is

$n_A= \frac{moles of A}{total number of moles} \\n_A=\frac{1}{4}$

$n_B$ moles of liquid B is

$n_B=\frac{moles of B}{ total number of moles} \\n_B=\frac{3}{4}$

Then you find the individual pressures, P_A and P_B, by multiplying initial pressures with mole fractions

$P_A= n_A \times P^{0}_A\\P_A=\frac{1}{4} \times 100mmHg\\P_A= 25mmHg$

$P_B= n_B \times P^{0} _B\\P_B= \frac{3}{4}  \times 80mmHg\\P_B= 60mmHg$

Then you add both to find the total pressure of the mixture

$P_T= P_A+P_B\\P_T=25+60\\P_T=85mmHg$

where P_T is the total pressure