Question

In Ostwald-Walker method, dry air is passed through a solution beaker, then a pure solvent beaker and then, through the dehydrating agent. Which of the following statement is 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. $\text{Loss in weight of solvent beaker}\propto p^{\circ }-p_s$
• B. $\text{Gain in weight of dehydrating agent}\propto p^{\circ }+p_s$
• C. $\text{Loss in weight of solvent beaker}\propto p^{\circ }$
• D. $\text{Loss in weight of solution beaker}\propto p^{\circ }-p_s$

Answer 1

The correct option is A $\text{Loss in weight of solvent beaker}\propto p^{\circ }-p_s$

When a dry air is passed through a solution beaker, 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.
$\therefore$
$\text{Loss in weight of solution beaker}\propto p_s$

When the air from solution beaker enters the solvent beaker some of the vapours of solvent gets along with the air and leaves the beaker because vapour pressure of solution will always be less than the vapour pressure of pure solvent . Again some solvent will vapourise. Thus, the loss in weight of solvent beaker depends on the difference in vapour pressure of pure solvent and solution.
$\therefore$
$\text{Loss in weight of solvent beaker}\propto p^{\circ }-p_s$

Finally, when the air is pass through the dehydrating agent, the vapours from the solvent is absorbed by dehydrating agent and result in gain in weight of dehydration agent.
Thus, its weight gain is directly proportional to the vapour pressure of pure solvent.

$\therefore \text{Gain in weight of dehydrating agent}\propto p^{\circ }$

Thus, statement (a) is correct.