Question

At 363 K, the vapour pressure of A is 21 kPa and that of B is 18 kPa. One mole of A and 2 moles of B are mixed. Assuming that this solution is ideal, the vapour pressure of the mixture is_____kPa. (Round off to the nearest integer).

Answer 1

Hint: Law of partial pressure given by Dalton states that in a gaseous mixture, total pressure is the summation of partial pressure of gases individually.
\[{P_T} = {P_1} + {P_2} + {P_3} + .....\]

Formula used:
The Raoult’s law is,
\[{P_{solution}}\, = \,{X_{solvent}} \times \,{P^0}_{solvent}\]
Here, \[{X_{solvent}}\] is mole fraction and \[{P^0}_{solvent}\] is pure solvent's vapour pressure.

Complete Step by Step Solution:
Here, the gaseous mixture is composed of B and A gases. The gas A's vapour pressure is \[21\,{\rm{kPa}}\] and the gas B's vapour pressure is 18 kPa. Now, we have to calculate A and B's mole fractions.
Moles of A = 1
Moles of B = 2
\[ \Rightarrow {X_A} = \dfrac{1}{{1 + 2}} = \dfrac{1}{3}\]
So, the mole fraction of A is 1/3.

Now, we have to calculate the mole fraction B using the formula \[{X_A} + {X_B} = 1\] .
\[ \Rightarrow {X_B} = 1 - \dfrac{1}{3} = \dfrac{2}{3}\]
Now, we have to find out the pressure in total using the partial pressure law given by Dalton.
\[{P_T} = \,{X_A}{P^0}_A\, + \,\,{X_B}{P^0}_B\]
Now, we have to put the value of mole fraction and partial vapour pressure of A and B in the above equation.
\[{P_T} = \,\,\dfrac{1}{3} \times 21 + \,\,\dfrac{2}{3} \times 18\]
\[{P_T} = 7 + 12 = 19\,\,{\rm{kPa}}\]
Hence, the total pressure of mixture is 19 kPa.

Note: The phenomenon of vapour pressure lowering states that addition of a non-volatile solute to a solvent results in the decrease of vapour pressure than the pure solvent. Therefore, one example of a colligative property is vapour pressure.