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Two liquids of vapour pressure p1 and p2 in pure state in the ratio of 2:1 are mixed in the molar ratio of 1:2. The ratio of their moles in the vapour state would be A) 1:1 B) 1:2 C) 2:1 D) 3:2?
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Two liquids of vapour pressure p1 and p2 in pure state in the ratio of...
Solution:
Given,
Liquids with vapour pressure p1 and p2 in pure state in the ratio of 2:1.
Molar ratio of liquids in the mixture = 1:2.
To find: The ratio of their moles in the vapour state.
Let us assume that we have '2x' moles of liquid 1 and 'x' moles of liquid 2 in pure state.
In the vapour state, the partial pressure of liquid 1 (p1) = mole fraction of liquid 1 in vapour state × total pressure.
Similarly, the partial pressure of liquid 2 (p2) = mole fraction of liquid 2 in vapour state × total pressure.
From Raoult's law, we know that mole fraction of liquid 1 in vapour state = moles of liquid 1 in vapour state / total moles in vapour state.
Total moles in vapour state = 2x + x = 3x.
Moles of liquid 1 in vapour state = (molar ratio of liquid 1 in the mixture) × (total moles in vapour state) = (1/3) × 3x = x.
Similarly, moles of liquid 2 in vapour state = (molar ratio of liquid 2 in the mixture) × (total moles in vapour state) = (2/3) × 3x = 2x.
Therefore, the ratio of their moles in the vapour state = x : 2x = 1:2.
Hence, the correct option is (B) 1:2.
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Solution:
Given,
Liquids with vapour pressure p1 and p2 in pure state in the ratio of 2:1.
Molar ratio of liquids in the mixture = 1:2.
To find: The ratio of their moles in the vapour state.
Assumptions:
Let us assume that we have '2x' moles of liquid 1 and 'x' moles of liquid 2 in pure state.
Calculation:
Partial pressure of liquid 1 (p1) = mole fraction of liquid 1 in vapour state × total pressure.
Partial pressure of liquid 2 (p2) = mole fraction of liquid 2 in vapour state × total pressure.
Mole fraction of liquid 1 in vapour state = moles of liquid 1 in vapour state / total moles in vapour state.
Total moles in vapour state = 2x + x = 3x.
Moles of liquid 1 in vapour state = (molar ratio of liquid 1 in the mixture) × (total moles in vapour state) = (1/3) × 3x = x.
Moles of liquid 2 in vapour state = (molar ratio of liquid 2 in the mixture) × (total moles in vapour state) = (2/3) × 3x = 2x.
Answer:
Ratio of their moles in the vapour state = x : 2x = 1:2.
Hence, the correct option is
Given,
Liquids with vapour pressure p1 and p2 in pure state in the ratio of 2:1.
Molar ratio of liquids in the mixture = 1:2.
To find: The ratio of their moles in the vapour state.
Let us assume that we have '2x' moles of liquid 1 and 'x' moles of liquid 2 in pure state.
In the vapour state, the partial pressure of liquid 1 (p1) = mole fraction of liquid 1 in vapour state × total pressure.
Similarly, the partial pressure of liquid 2 (p2) = mole fraction of liquid 2 in vapour state × total pressure.
From Raoult's law, we know that mole fraction of liquid 1 in vapour state = moles of liquid 1 in vapour state / total moles in vapour state.
Total moles in vapour state = 2x + x = 3x.
Moles of liquid 1 in vapour state = (molar ratio of liquid 1 in the mixture) × (total moles in vapour state) = (1/3) × 3x = x.
Similarly, moles of liquid 2 in vapour state = (molar ratio of liquid 2 in the mixture) × (total moles in vapour state) = (2/3) × 3x = 2x.
Therefore, the ratio of their moles in the vapour state = x : 2x = 1:2.
Hence, the correct option is (B) 1:2.
HTML Code:
Solution:
Given,
Liquids with vapour pressure p1 and p2 in pure state in the ratio of 2:1.
Molar ratio of liquids in the mixture = 1:2.
To find: The ratio of their moles in the vapour state.
Assumptions:
Let us assume that we have '2x' moles of liquid 1 and 'x' moles of liquid 2 in pure state.
Calculation:
Partial pressure of liquid 1 (p1) = mole fraction of liquid 1 in vapour state × total pressure.
Partial pressure of liquid 2 (p2) = mole fraction of liquid 2 in vapour state × total pressure.
Mole fraction of liquid 1 in vapour state = moles of liquid 1 in vapour state / total moles in vapour state.
Total moles in vapour state = 2x + x = 3x.
Moles of liquid 1 in vapour state = (molar ratio of liquid 1 in the mixture) × (total moles in vapour state) = (1/3) × 3x = x.
Moles of liquid 2 in vapour state = (molar ratio of liquid 2 in the mixture) × (total moles in vapour state) = (2/3) × 3x = 2x.
Answer:
Ratio of their moles in the vapour state = x : 2x = 1:2.
Hence, the correct option is
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Two liquids of vapour pressure p1 and p2 in pure state in the ratio of 2:1 are mixed in the molar ratio of 1:2. The ratio of their moles in the vapour state would be A) 1:1 B) 1:2 C) 2:1 D) 3:2?
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Two liquids of vapour pressure p1 and p2 in pure state in the ratio of 2:1 are mixed in the molar ratio of 1:2. The ratio of their moles in the vapour state would be A) 1:1 B) 1:2 C) 2:1 D) 3:2? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two liquids of vapour pressure p1 and p2 in pure state in the ratio of 2:1 are mixed in the molar ratio of 1:2. The ratio of their moles in the vapour state would be A) 1:1 B) 1:2 C) 2:1 D) 3:2? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two liquids of vapour pressure p1 and p2 in pure state in the ratio of 2:1 are mixed in the molar ratio of 1:2. The ratio of their moles in the vapour state would be A) 1:1 B) 1:2 C) 2:1 D) 3:2?.
Two liquids of vapour pressure p1 and p2 in pure state in the ratio of 2:1 are mixed in the molar ratio of 1:2. The ratio of their moles in the vapour state would be A) 1:1 B) 1:2 C) 2:1 D) 3:2? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two liquids of vapour pressure p1 and p2 in pure state in the ratio of 2:1 are mixed in the molar ratio of 1:2. The ratio of their moles in the vapour state would be A) 1:1 B) 1:2 C) 2:1 D) 3:2? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two liquids of vapour pressure p1 and p2 in pure state in the ratio of 2:1 are mixed in the molar ratio of 1:2. The ratio of their moles in the vapour state would be A) 1:1 B) 1:2 C) 2:1 D) 3:2?.
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Two liquids of vapour pressure p1 and p2 in pure state in the ratio of 2:1 are mixed in the molar ratio of 1:2. The ratio of their moles in the vapour state would be A) 1:1 B) 1:2 C) 2:1 D) 3:2?, a detailed solution for Two liquids of vapour pressure p1 and p2 in pure state in the ratio of 2:1 are mixed in the molar ratio of 1:2. The ratio of their moles in the vapour state would be A) 1:1 B) 1:2 C) 2:1 D) 3:2? has been provided alongside types of Two liquids of vapour pressure p1 and p2 in pure state in the ratio of 2:1 are mixed in the molar ratio of 1:2. The ratio of their moles in the vapour state would be A) 1:1 B) 1:2 C) 2:1 D) 3:2? theory, EduRev gives you an
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