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When a liquid that is immiscible with water was steam distilled at 95.2°C at a total pressure of 748 torr, the distillate contained 1.25 g of the liquid per gram of water. The vapour pressure of water is 648 torr at 95.2°C, what is the molar mass of liquid?
- a)7.975 g/mol.
- b)166 g/mol.
- c)145.8 g/mol.
- d)None of these.
Correct answer is option 'C'. Can you explain this answer?
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When a liquid that is immiscible with water was steam distilled at 95....
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When a liquid that is immiscible with water was steam distilled at 95....
Given information:
- A liquid that is immiscible with water was steam distilled at 95.2C at a total pressure of 748 torr.
- The distillate contained 1.25 g of the liquid per gram of water.
- The vapour pressure of water is 648 torr at 95.2C.
To find: the molar mass of the liquid.
Solution:
1. Calculate the mole fraction of the liquid in the distillate:
Mole fraction of liquid = mass of liquid / (mass of liquid + mass of water)
Given, the distillate contained 1.25 g of the liquid per gram of water.
Let's assume we take 100 g of the distillate, then it will contain:
- 1.25 g of the liquid
- 100 g of water
Therefore, the mole fraction of the liquid in the distillate = 1.25 / (1.25 + 100) = 0.01235
2. Calculate the mole fraction of water in the distillate:
Mole fraction of water = mass of water / (mass of liquid + mass of water)
Using the same assumption as before, the mass of water in 100 g of the distillate = 100 g.
Therefore, the mole fraction of water in the distillate = 100 / (1.25 + 100) = 0.9876
3. Calculate the mole fraction of water in the vapour phase:
Using Dalton's law of partial pressures, the total pressure (P) is the sum of the partial pressures of water and the liquid:
P = P_water + P_liquid
where,
P_water = vapour pressure of water = 648 torr
P_liquid = partial pressure of the liquid
Total pressure = 748 torr (given)
Therefore,
P_liquid = P - P_water = 748 - 648 = 100 torr
The mole fraction of water in the vapour phase can be calculated using Raoult's law:
P_water = X_water * P_total
where,
X_water = mole fraction of water in the vapour phase (to be calculated)
Substituting the given values:
648 = X_water * 748
X_water = 648 / 748 = 0.8661
Therefore, the mole fraction of the liquid in the vapour phase = 1 - 0.8661 = 0.1339
4. Calculate the molar mass of the liquid:
Let's assume the molar mass of the liquid = M.
Using the mole fraction of the liquid in the distillate and the vapour phase, we can write:
Mole fraction of liquid in distillate = X_liquid * P_total / P_liquid
0.01235 = X_liquid * 748 / 100
X_liquid = 0.01235 * 100 / 748 = 0.00165
Mole fraction of liquid in vapour phase = X_liquid * P_total / P_liquid
0.1339 = X_liquid * 748 / 100
X_liquid = 0.1339 * 100 / 748 = 0.0179
Now, using the ratio of mole fractions in the distillate and the vapour phase, we can write:
0.00165
- A liquid that is immiscible with water was steam distilled at 95.2C at a total pressure of 748 torr.
- The distillate contained 1.25 g of the liquid per gram of water.
- The vapour pressure of water is 648 torr at 95.2C.
To find: the molar mass of the liquid.
Solution:
1. Calculate the mole fraction of the liquid in the distillate:
Mole fraction of liquid = mass of liquid / (mass of liquid + mass of water)
Given, the distillate contained 1.25 g of the liquid per gram of water.
Let's assume we take 100 g of the distillate, then it will contain:
- 1.25 g of the liquid
- 100 g of water
Therefore, the mole fraction of the liquid in the distillate = 1.25 / (1.25 + 100) = 0.01235
2. Calculate the mole fraction of water in the distillate:
Mole fraction of water = mass of water / (mass of liquid + mass of water)
Using the same assumption as before, the mass of water in 100 g of the distillate = 100 g.
Therefore, the mole fraction of water in the distillate = 100 / (1.25 + 100) = 0.9876
3. Calculate the mole fraction of water in the vapour phase:
Using Dalton's law of partial pressures, the total pressure (P) is the sum of the partial pressures of water and the liquid:
P = P_water + P_liquid
where,
P_water = vapour pressure of water = 648 torr
P_liquid = partial pressure of the liquid
Total pressure = 748 torr (given)
Therefore,
P_liquid = P - P_water = 748 - 648 = 100 torr
The mole fraction of water in the vapour phase can be calculated using Raoult's law:
P_water = X_water * P_total
where,
X_water = mole fraction of water in the vapour phase (to be calculated)
Substituting the given values:
648 = X_water * 748
X_water = 648 / 748 = 0.8661
Therefore, the mole fraction of the liquid in the vapour phase = 1 - 0.8661 = 0.1339
4. Calculate the molar mass of the liquid:
Let's assume the molar mass of the liquid = M.
Using the mole fraction of the liquid in the distillate and the vapour phase, we can write:
Mole fraction of liquid in distillate = X_liquid * P_total / P_liquid
0.01235 = X_liquid * 748 / 100
X_liquid = 0.01235 * 100 / 748 = 0.00165
Mole fraction of liquid in vapour phase = X_liquid * P_total / P_liquid
0.1339 = X_liquid * 748 / 100
X_liquid = 0.1339 * 100 / 748 = 0.0179
Now, using the ratio of mole fractions in the distillate and the vapour phase, we can write:
0.00165
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When a liquid that is immiscible with water was steam distilled at 95.2°C at a total pressure of 748 torr, the distillate contained 1.25 g of the liquid per gram of water. The vapour pressure of water is 648 torr at 95.2°C, what is the molar mass of liquid?a)7.975 g/mol.b)166 g/mol.c)145.8 g/mol.d)None of these.Correct answer is option 'C'. Can you explain this answer?
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When a liquid that is immiscible with water was steam distilled at 95.2°C at a total pressure of 748 torr, the distillate contained 1.25 g of the liquid per gram of water. The vapour pressure of water is 648 torr at 95.2°C, what is the molar mass of liquid?a)7.975 g/mol.b)166 g/mol.c)145.8 g/mol.d)None of these.Correct answer is option 'C'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about When a liquid that is immiscible with water was steam distilled at 95.2°C at a total pressure of 748 torr, the distillate contained 1.25 g of the liquid per gram of water. The vapour pressure of water is 648 torr at 95.2°C, what is the molar mass of liquid?a)7.975 g/mol.b)166 g/mol.c)145.8 g/mol.d)None of these.Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When a liquid that is immiscible with water was steam distilled at 95.2°C at a total pressure of 748 torr, the distillate contained 1.25 g of the liquid per gram of water. The vapour pressure of water is 648 torr at 95.2°C, what is the molar mass of liquid?a)7.975 g/mol.b)166 g/mol.c)145.8 g/mol.d)None of these.Correct answer is option 'C'. Can you explain this answer?.
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