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The vapour pressure of a solvent decreased by 10mm of mercury when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be 20mm of mercury?
- a)0.8
- b)0.6 [1998]
- c)0.4
- d)0.2
Correct answer is option 'B'. Can you explain this answer?
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To solve this question, we need to use Raoult's law, which states that the vapor pressure of a solvent in a solution is directly proportional to its mole fraction in the solution. According to Raoult's law, the decrease in vapor pressure of the solvent is directly proportional to the mole fraction of the solute in the solution.
Let's assume that the initial vapor pressure of the solvent is P₀ and the final vapor pressure is P. We are given that the decrease in vapor pressure is 10 mm Hg when the mole fraction of the solute is 0.2. We need to find the mole fraction of the solvent when the decrease in vapor pressure is 20 mm Hg.
Let's calculate the initial mole fraction of the solute using the given information:
ΔP = P₀ - P = 10 mm Hg
X₁ = 0.2
Now, let's calculate the final mole fraction of the solvent when the decrease in vapor pressure is 20 mm Hg:
ΔP = P₀ - P = 20 mm Hg
X₂ = ?
According to Raoult's law, the decrease in vapor pressure is directly proportional to the mole fraction of the solute:
ΔP ∝ X
Using this proportionality, we can write the following equation:
ΔP₁/X₁ = ΔP₂/X₂
Substituting the given values, we get:
10/0.2 = 20/X₂
Now, let's solve for X₂:
X₂ = (20 * 0.2) / 10
X₂ = 0.4
Therefore, the mole fraction of the solvent when the decrease in vapor pressure is 20 mm Hg is 0.4.
Let's assume that the initial vapor pressure of the solvent is P₀ and the final vapor pressure is P. We are given that the decrease in vapor pressure is 10 mm Hg when the mole fraction of the solute is 0.2. We need to find the mole fraction of the solvent when the decrease in vapor pressure is 20 mm Hg.
Let's calculate the initial mole fraction of the solute using the given information:
ΔP = P₀ - P = 10 mm Hg
X₁ = 0.2
Now, let's calculate the final mole fraction of the solvent when the decrease in vapor pressure is 20 mm Hg:
ΔP = P₀ - P = 20 mm Hg
X₂ = ?
According to Raoult's law, the decrease in vapor pressure is directly proportional to the mole fraction of the solute:
ΔP ∝ X
Using this proportionality, we can write the following equation:
ΔP₁/X₁ = ΔP₂/X₂
Substituting the given values, we get:
10/0.2 = 20/X₂
Now, let's solve for X₂:
X₂ = (20 * 0.2) / 10
X₂ = 0.4
Therefore, the mole fraction of the solvent when the decrease in vapor pressure is 20 mm Hg is 0.4.
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The vapour pressure of a solvent decreased by 10mm of mercury when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be 20mm of mercury?a)0.8b)0.6 [1998]c)0.4d)0.2Correct answer is option 'B'. Can you explain this answer?
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The vapour pressure of a solvent decreased by 10mm of mercury when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be 20mm of mercury?a)0.8b)0.6 [1998]c)0.4d)0.2Correct answer is option 'B'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The vapour pressure of a solvent decreased by 10mm of mercury when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be 20mm of mercury?a)0.8b)0.6 [1998]c)0.4d)0.2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The vapour pressure of a solvent decreased by 10mm of mercury when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be 20mm of mercury?a)0.8b)0.6 [1998]c)0.4d)0.2Correct answer is option 'B'. Can you explain this answer?.
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