Question

In Ostwald-Walker method, dry air is passed through a series of beaker containing solution and solvent and finally through a dehydrating agent as shown in the figure
Which of the following statement(s) is/are correct regarding the above setup?
Here,
$p^{\circ }\text{is the vapour pressure of the pure solvent}$
$p_s\text{is the vapour pressure of the solution}$

• A. Weight of beaker 3 increase
• B. Weight of beaker 2 increases
• C. Weight of beaker 4 decreases
• D. $\text{Loss in weight of beaker 1}\propto p_s$
  

Answer 1

Answer: (A), (C), (D)

$\text{Loss in weight of beaker 1}\propto p_s$


Beaker 1:
When a dry air is passed through a solution beaker '1', the dry air will get some vapours of the solution and leaves the beaker. Due to loss of vapour pressure of solution, more amount of solvent vapourise to nullify this effect. Thus, the weight of the solution will decrease and is directly proportional to the vapour pressure of the solution.
$\therefore$
$\text{Loss in weight of solution beaker}\propto p_s$
Thus, statement (d) is correct.

Beaker 2:
When the wet air from beaker '1' enters the solvent beaker '2' some of the vapours of solvent gets along with the air and leaves the beaker '2' because vapour pressure of solution will always be less than the vapour pressure of pure solvent . So, again some solvent will vapourise. Thus, there will be loss in weight of solvent beaker. The weight loss depends on the difference in vapour pressure of pure solvent and solution.
$\therefore$
$\text{Loss in weight of solvent beaker}\propto p^{\circ }-p_s$
Thus, statement (b) is wrong.

Beaker 3:
When the wet air enters solution beaker '3', since the vapour pressure of solution is less than solvent it will gain some vapour from the air and to nullify the effect some vapour will condense and increase the weight of the solution. Thus, weight of the beaker increase and it depends on the difference in vapour pressure of pure solvent and solution.
Thus, statement (a) is correct.

Beaker 4:
When the wet air from beaker '3' enters the solvent beaker '4' some of the vapours of solvent gets along with the air because vapour pressure of solution will always be less than the vapour pressure of pure solvent. So, again some solvent will vapourise. Thus, there will be loss in weight of solvent beaker. The weight loss depends on the difference in vapour pressure of pure solvent and solution.
Thus, statement (c) is correct.

In beaker 5, there will increase in weight because it will absorbs the vapours from the air and the weight gain is directly proportional to the pure vapour pressure of solvent.