Question

The vapour pressure of two liquids P and Q are \[\text{80 torr}\]and \[\text{60 torr}\] respectively. The total vapour pressure obtained by mixing 3 mole of P and 2 mole of Q would be:
A.\[\text{68 torr}\]
B.\[\text{20 torr}\]
C.\[\text{140 torr}\]
D.\[\text{72 torr}\]

Answer 1

Hint: The vapour pressure liquids are related to their mole fraction in the solution by the relation given by the Raoult’s Law which states that the vapour pressure of a solution is equal to the product of the vapour pressure of the solution multiplied by the mole fraction of the solute in the solution.

Complete step by step answer:
The vapour pressure of two liquids P and Q are \[\text{80 torr}\] and \[\text{60 torr}\] respectively. The moles of the liquids are: 3 mole of P and 2 mole of Q.
The mole-fraction of P = $\dfrac{3}{3+2}=\dfrac{3}{5}$and that for Q is, $\dfrac{2}{3+2}=\dfrac{2}{5}$
Now the total vapour pressure of the solution is,
(Mole fraction of P x the vapour pressure of P) + (the mole fraction of Q x the vapour pressure of Q) = $\left( \dfrac{3}{5}\times 80 \right)\text{ }+\text{ }\left( \dfrac{2}{5}\times 60 \right)=\text{ }48+24\text{ }=72\text{ torr}$
The total vapour pressure obtained by mixing 3 mole of P and 2 mole of Q would be \[\text{72 torr}\].

So, the correct answer is option D.

Note:
In this regard, the point that is worth-mentioning is that the Raoult’s Law works only in case of ideal solutions only, where an ideal solution refers to those solutions in which there is no association or dissociation of the solute molecules. For example, sodium chloride dissociates into ions in the aqueous solution and hence it behaves as a non-ideal solution. Same is the case for acetic acid that associates in solution forming dimers and hence it is also a non-ideal solution.