1
Question
Consider the following cases A and B in Ostwald-Walker Method:
Case A : Dry air is passed through a pure solvent and then, a dehydrating agent.
Case B: Dry air is passed through a pure solvent and then a solution with non-volatile solute and then, a dehydrating agent
Here,
p∘ is the vapour pressure of the pure solvent
ps is the vapour pressure of the solution
Choose the incorrect option:
Consider the following cases A and B in Ostwald-Walker Method:
Case A : Dry air is passed through a pure solvent and then, a dehydrating agent.
Case B: Dry air is passed through a pure solvent and then a solution with non-volatile solute and then, a dehydrating agent
Here,
p∘ is the vapour pressure of the pure solvent
ps is the vapour pressure of the solution
Choose the incorrect option:
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Solution
The correct option is C In case B, loss of weight of solvent container ∝ PS
Case A : Dry air is passed through a pure solvent and then, a dehydrating agent.
As the vapour pressure of the solvent increases, more are the vapours formed, and vapours that the dry air carries is proportional to the vapour pressure of pure solvent. The weight of the solvent decreases because it re tries to establish the equilibrium which is disturbed by absorption of vapours by dry air. Absorption of vapours of dry air is proportional to the vapour pressure of pure solvent which in turn is proportional to loss of weight of solvent
So (b) is correct.
The dehydrating agent will gain weight from those vapours which are absorbed by dry air. As the vapour pressure of pure solvent increases, more are the vapours absorbed by dry air and more is the gain in weight of dehydrating agent.
So (a) is correct.
Case B: Dry air is passed through a pure solvent and then a solution with non-volatile solute and then, a dehydrating agent
More the vapour pressure of solvent, more will be the vapours absorbed by dry air, and in turn the solvent will try to reestablish the equilibrium, for those absorbed vapours.
So, loss of weight of solvent is proportional to vapour pressure of pure solvent, not solution.
Hence, (c) is incorrect.
After the dry air is passed through solvent it is passed through solution containing non-volatile solute
In solution containing non volatile solute, the vapour pressure is lowered. So, less vapours are formed. Now the equilibrium will shift from dry air to the the solution containing non-volatile solute and a new equilibrium is established.
So, the gain in weight of solution ∝ (P∘−PS)
Also, here
Gain in weight of dehydrating agent ∝PS
So (d) is correct.
Case A : Dry air is passed through a pure solvent and then, a dehydrating agent.
As the vapour pressure of the solvent increases, more are the vapours formed, and vapours that the dry air carries is proportional to the vapour pressure of pure solvent. The weight of the solvent decreases because it re tries to establish the equilibrium which is disturbed by absorption of vapours by dry air. Absorption of vapours of dry air is proportional to the vapour pressure of pure solvent which in turn is proportional to loss of weight of solvent
So (b) is correct.
The dehydrating agent will gain weight from those vapours which are absorbed by dry air. As the vapour pressure of pure solvent increases, more are the vapours absorbed by dry air and more is the gain in weight of dehydrating agent.
So (a) is correct.
Case B: Dry air is passed through a pure solvent and then a solution with non-volatile solute and then, a dehydrating agent
More the vapour pressure of solvent, more will be the vapours absorbed by dry air, and in turn the solvent will try to reestablish the equilibrium, for those absorbed vapours.
So, loss of weight of solvent is proportional to vapour pressure of pure solvent, not solution.
Hence, (c) is incorrect.
After the dry air is passed through solvent it is passed through solution containing non-volatile solute
In solution containing non volatile solute, the vapour pressure is lowered. So, less vapours are formed. Now the equilibrium will shift from dry air to the the solution containing non-volatile solute and a new equilibrium is established.
So, the gain in weight of solution ∝ (P∘−PS)
Also, here
Gain in weight of dehydrating agent ∝PS
So (d) is correct.
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