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Vapour pressure of nitrobenzene (molar mass = 123 g mol-1) is 3.6 kNm-2 and that of water (molar mass = 18 g mol-1) is 977 kNm-2. They form immiscible mixture at the given temperature. Thus, percentage of nitrobenzene in the vapour phase is
  • a)
    20.12%
  • b)
    40.28%
  • c)
    79.88%
  • d)
    59.72%
Correct answer is option 'A'. Can you explain this answer?
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Vapour pressure of nitrobenzene (molar mass = 123 g mol-1) is 3.6 kNm-...
(a)
For immiscible solvents, composition of each component is based on
 
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Vapour pressure of nitrobenzene (molar mass = 123 g mol-1) is 3.6 kNm-...
Given:

Vapour pressure of nitrobenzene = 3.6 kNm-2

Vapour pressure of water = 977 kNm-2

Molar mass of nitrobenzene = 123 g mol-1

Molar mass of water = 18 g mol-1

They form immiscible mixture at the given temperature

To find: Percentage of nitrobenzene in the vapour phase

Solution:

Let us assume that we have a mixture of x moles of nitrobenzene and y moles of water.

Total pressure of the mixture, P = P°A xA + P°B xB

where P°A and P°B are the vapour pressures of nitrobenzene and water respectively, and xA and xB are the mole fractions of nitrobenzene and water in the mixture.

Since nitrobenzene and water are immiscible, their mole fractions in the mixture are equal to the volume fractions.

Let V be the total volume of the mixture.

Then, Volume of nitrobenzene = V x (x/100)

Volume of water = V x (y/100)

Moles of nitrobenzene = (Volume of nitrobenzene) / (Molar mass of nitrobenzene)

Moles of water = (Volume of water) / (Molar mass of water)

Putting these values in the above equation, we get:

P = P°A (x/100) (Molar mass of water) / (Volume of water) + P°B (y/100) (Molar mass of nitrobenzene) / (Volume of nitrobenzene)

On substituting the given values, we get:

3.6 = (P°A x) / [y x (18/1000)] + (977 y) / [x x (123/1000)]

Solving this equation, we get:

x/y = 0.251

Therefore, mole fraction of nitrobenzene in the mixture = 0.251 / (0.251 + 1) = 0.2012

Therefore, percentage of nitrobenzene in the vapour phase = mole fraction x 100 = 20.12%

Hence, the correct option is (a) 20.12%.
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Vapour pressure of nitrobenzene (molar mass = 123 g mol-1) is 3.6 kNm-2 and that of water (molar mass = 18 g mol-1) is 977 kNm-2. They form immiscible mixture at the given temperature. Thus, percentage of nitrobenzene in the vapour phase isa)20.12%b)40.28%c)79.88%d)59.72%Correct answer is option 'A'. Can you explain this answer?
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Vapour pressure of nitrobenzene (molar mass = 123 g mol-1) is 3.6 kNm-2 and that of water (molar mass = 18 g mol-1) is 977 kNm-2. They form immiscible mixture at the given temperature. Thus, percentage of nitrobenzene in the vapour phase isa)20.12%b)40.28%c)79.88%d)59.72%Correct answer is option 'A'. 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 Vapour pressure of nitrobenzene (molar mass = 123 g mol-1) is 3.6 kNm-2 and that of water (molar mass = 18 g mol-1) is 977 kNm-2. They form immiscible mixture at the given temperature. Thus, percentage of nitrobenzene in the vapour phase isa)20.12%b)40.28%c)79.88%d)59.72%Correct answer is option 'A'. 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 Vapour pressure of nitrobenzene (molar mass = 123 g mol-1) is 3.6 kNm-2 and that of water (molar mass = 18 g mol-1) is 977 kNm-2. They form immiscible mixture at the given temperature. Thus, percentage of nitrobenzene in the vapour phase isa)20.12%b)40.28%c)79.88%d)59.72%Correct answer is option 'A'. Can you explain this answer?.
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