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The solubility of a specific non–volatile salt is 4 g in 100 g of water at 25°C. If 2.0 g, 4.0 g and 6.0 g of the salt added of 100 g of water at 25°C, in system X, Y, and Z. The vapour pressure would be in the order:
- a)X < Y < Z
- b)X > Y > Z
- c)Z > X = Y
- d)X > Y = Z
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
The solubility of a specific non–volatile salt is 4 g in 100 g o...
Solution X is unsaturated so vapour pressure will be more, solution Y & Z are saturated so vapour pressure of Y = vapour pressure of Z and 2 gm of solute would be present in form of solid in system Z.
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The solubility of a specific non–volatile salt is 4 g in 100 g o...
Given information:
- Solubility of a nonvolatile salt = 4 g in 100 g of water at 25°C
- Mass of salt added to 100 g of water: 2.0 g, 4.0 g, and 6.0 g
- Temperature = 25°C
To determine the order of vapor pressure in the systems X, Y, and Z, we need to understand the effect of solute concentration on vapor pressure.
Effect of solute concentration on vapor pressure:
- When a nonvolatile solute is added to a solvent, it lowers the vapor pressure of the solvent.
- This is because the solute molecules occupy some of the surface area of the solvent, reducing the number of solvent molecules that can escape into the gas phase.
- As a result, the vapor pressure of the solvent is reduced.
Order of vapor pressure in the systems X, Y, and Z:
- System X: 2.0 g of salt added to 100 g of water
- System Y: 4.0 g of salt added to 100 g of water
- System Z: 6.0 g of salt added to 100 g of water
To determine the order of vapor pressure, we need to calculate the molality of each solution and use it to calculate the vapor pressure using Raoult's law.
Molality of the solutions:
- Molality (m) = moles of solute / mass of solvent (in kg)
- Mass of solvent (in kg) = 100 g / 1000 = 0.1 kg
- Moles of solute:
- System X: 2.0 g / 58.44 g/mol = 0.034 mol
- System Y: 4.0 g / 58.44 g/mol = 0.068 mol
- System Z: 6.0 g / 58.44 g/mol = 0.102 mol
- Molality:
- System X: 0.034 mol / 0.1 kg = 0.34 m
- System Y: 0.068 mol / 0.1 kg = 0.68 m
- System Z: 0.102 mol / 0.1 kg = 1.02 m
Vapor pressure of the solutions:
- Vapor pressure (P) = mole fraction of solvent * vapor pressure of pure solvent
- Mole fraction of solvent (Xs) = 1 - mole fraction of solute
- Vapor pressure of pure solvent (Po) = 23.76 mmHg (at 25°C)
- Mole fraction of solute:
- System X: 0.034 / (0.034 + 0.998) = 0.033
- System Y: 0.068 / (0.068 + 0.932) = 0.068
- System Z: 0.102 / (0.102 + 0.898) = 0.102
- Mole fraction of solvent:
- System X: 1 - 0.033 = 0.967
- System Y: 1 - 0.068 = 0.932
- System Z: 1 - 0.102 = 0.898
- Vapor pressure:
-
- Solubility of a nonvolatile salt = 4 g in 100 g of water at 25°C
- Mass of salt added to 100 g of water: 2.0 g, 4.0 g, and 6.0 g
- Temperature = 25°C
To determine the order of vapor pressure in the systems X, Y, and Z, we need to understand the effect of solute concentration on vapor pressure.
Effect of solute concentration on vapor pressure:
- When a nonvolatile solute is added to a solvent, it lowers the vapor pressure of the solvent.
- This is because the solute molecules occupy some of the surface area of the solvent, reducing the number of solvent molecules that can escape into the gas phase.
- As a result, the vapor pressure of the solvent is reduced.
Order of vapor pressure in the systems X, Y, and Z:
- System X: 2.0 g of salt added to 100 g of water
- System Y: 4.0 g of salt added to 100 g of water
- System Z: 6.0 g of salt added to 100 g of water
To determine the order of vapor pressure, we need to calculate the molality of each solution and use it to calculate the vapor pressure using Raoult's law.
Molality of the solutions:
- Molality (m) = moles of solute / mass of solvent (in kg)
- Mass of solvent (in kg) = 100 g / 1000 = 0.1 kg
- Moles of solute:
- System X: 2.0 g / 58.44 g/mol = 0.034 mol
- System Y: 4.0 g / 58.44 g/mol = 0.068 mol
- System Z: 6.0 g / 58.44 g/mol = 0.102 mol
- Molality:
- System X: 0.034 mol / 0.1 kg = 0.34 m
- System Y: 0.068 mol / 0.1 kg = 0.68 m
- System Z: 0.102 mol / 0.1 kg = 1.02 m
Vapor pressure of the solutions:
- Vapor pressure (P) = mole fraction of solvent * vapor pressure of pure solvent
- Mole fraction of solvent (Xs) = 1 - mole fraction of solute
- Vapor pressure of pure solvent (Po) = 23.76 mmHg (at 25°C)
- Mole fraction of solute:
- System X: 0.034 / (0.034 + 0.998) = 0.033
- System Y: 0.068 / (0.068 + 0.932) = 0.068
- System Z: 0.102 / (0.102 + 0.898) = 0.102
- Mole fraction of solvent:
- System X: 1 - 0.033 = 0.967
- System Y: 1 - 0.068 = 0.932
- System Z: 1 - 0.102 = 0.898
- Vapor pressure:
-
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The solubility of a specific non–volatile salt is 4 g in 100 g o...
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The solubility of a specific non–volatile salt is 4 g in 100 g of water at 25°C. If 2.0 g, 4.0 g and 6.0 g of the salt added of 100 g of water at 25°C, in system X, Y, and Z. The vapour pressure would be in the order:a)X < Y < Zb)X > Y > Zc)Z > X = Yd)X > Y = ZCorrect answer is option 'D'. Can you explain this answer? for Chemistry 2025 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about The solubility of a specific non–volatile salt is 4 g in 100 g of water at 25°C. If 2.0 g, 4.0 g and 6.0 g of the salt added of 100 g of water at 25°C, in system X, Y, and Z. The vapour pressure would be in the order:a)X < Y < Zb)X > Y > Zc)Z > X = Yd)X > Y = ZCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Chemistry 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The solubility of a specific non–volatile salt is 4 g in 100 g of water at 25°C. If 2.0 g, 4.0 g and 6.0 g of the salt added of 100 g of water at 25°C, in system X, Y, and Z. The vapour pressure would be in the order:a)X < Y < Zb)X > Y > Zc)Z > X = Yd)X > Y = ZCorrect answer is option 'D'. Can you explain this answer?.
The solubility of a specific non–volatile salt is 4 g in 100 g of water at 25°C. If 2.0 g, 4.0 g and 6.0 g of the salt added of 100 g of water at 25°C, in system X, Y, and Z. The vapour pressure would be in the order:a)X < Y < Zb)X > Y > Zc)Z > X = Yd)X > Y = ZCorrect answer is option 'D'. Can you explain this answer? for Chemistry 2025 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about The solubility of a specific non–volatile salt is 4 g in 100 g of water at 25°C. If 2.0 g, 4.0 g and 6.0 g of the salt added of 100 g of water at 25°C, in system X, Y, and Z. The vapour pressure would be in the order:a)X < Y < Zb)X > Y > Zc)Z > X = Yd)X > Y = ZCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Chemistry 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The solubility of a specific non–volatile salt is 4 g in 100 g of water at 25°C. If 2.0 g, 4.0 g and 6.0 g of the salt added of 100 g of water at 25°C, in system X, Y, and Z. The vapour pressure would be in the order:a)X < Y < Zb)X > Y > Zc)Z > X = Yd)X > Y = ZCorrect answer is option 'D'. Can you explain this answer?.
Solutions for The solubility of a specific non–volatile salt is 4 g in 100 g of water at 25°C. If 2.0 g, 4.0 g and 6.0 g of the salt added of 100 g of water at 25°C, in system X, Y, and Z. The vapour pressure would be in the order:a)X < Y < Zb)X > Y > Zc)Z > X = Yd)X > Y = ZCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry.
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