
The vapour pressure of pure water is 32.8mm of Hg. It is mixed with ethanol which has mole fraction 0.9. Find the vapour pressure of the resulting solution:
(A) 39.3mm Hg
(B) 36.0mm Hg
(C) 29.5mm Hg
(D) 28.8mm Hg
Answer
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Hint: Vapor pressure or we can say equilibrium vapour pressure is defined as the pressure exerted by a vapour in equilibrium with its condensed phases (either solid or liquid) at a given temperature in a closed system.
Complete step by step solution:
First of all, let’s check how this vapour pressure changes, or more precisely vapour pressure depression occurs.
When any solute is added to the solvent, space near the surface of the liquid is occupied by some of the solute molecules. When a solute is dissolved in a solvent, the number of solvent molecules near the surface changes, and the vapour pressure of the solvent changes. This has no effect on the rate at which solvent molecules in the gas phase condense to go back in it's the liquid state. But it decreases the rate at which the solvent in the liquid phase gets converted into gas. As a result, the vapour pressure of the solvent (escaping from a solution) should be smaller than the vapour pressure of the previously present solvent which was pure.
Now, since we know how it works, let’s jump right back to our question.
We know that the reduction of the vapour pressure of the solvent by adding a solute is given as follows:
${P_1} = x.{P_1}^0$
Vapour pressure of the solution = mole fraction of solute and Vapor pressure of the pure solvent
According to the question, vapour pressure of pure water = $32.8 mm of Hg$
Mole fraction of ethanol (x) = $0.9$
Now by substituting the values to the above equation, we get
Vapor pressure of solution (${p_1}$) = $0.9×32.8 = 29.52 mm of Hg$
Therefore, the correct answer is (C) 29.5mm Hg.
Note: The change in the vapour pressure occurs when a solute is added to a solvent. Therefore, it is a colligative property. If it depends on the mole fraction of the solute, then it must also depend on the ratio of the number of particles of solute to solvent in the solution but not the identity of the given solute.
Complete step by step solution:
First of all, let’s check how this vapour pressure changes, or more precisely vapour pressure depression occurs.
When any solute is added to the solvent, space near the surface of the liquid is occupied by some of the solute molecules. When a solute is dissolved in a solvent, the number of solvent molecules near the surface changes, and the vapour pressure of the solvent changes. This has no effect on the rate at which solvent molecules in the gas phase condense to go back in it's the liquid state. But it decreases the rate at which the solvent in the liquid phase gets converted into gas. As a result, the vapour pressure of the solvent (escaping from a solution) should be smaller than the vapour pressure of the previously present solvent which was pure.
Now, since we know how it works, let’s jump right back to our question.
We know that the reduction of the vapour pressure of the solvent by adding a solute is given as follows:
${P_1} = x.{P_1}^0$
Vapour pressure of the solution = mole fraction of solute and Vapor pressure of the pure solvent
According to the question, vapour pressure of pure water = $32.8 mm of Hg$
Mole fraction of ethanol (x) = $0.9$
Now by substituting the values to the above equation, we get
Vapor pressure of solution (${p_1}$) = $0.9×32.8 = 29.52 mm of Hg$
Therefore, the correct answer is (C) 29.5mm Hg.
Note: The change in the vapour pressure occurs when a solute is added to a solvent. Therefore, it is a colligative property. If it depends on the mole fraction of the solute, then it must also depend on the ratio of the number of particles of solute to solvent in the solution but not the identity of the given solute.
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