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A saturated aqueous solution of sparingly soluble salt AB3 has the vapour pressure 0.08 mm lesser than the vapour pressure 17.33 mm of solvent at 25oC. The solubility product of AB3 is :
- a)1.087 x 10-2
- b)1.48 x 10-4
- c)5.35 x 10-5
- d)4.56 x 10-4
Correct answer is option 'D'. Can you explain this answer?
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A saturated aqueous solution of sparingly soluble salt AB3has the vapo...
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A saturated aqueous solution of sparingly soluble salt AB3has the vapo...
Given:
- Vapour pressure of the solvent (P°) = 17.33 mm
- Vapour pressure of the saturated solution (P) = 17.33 - 0.08 = 17.25 mm
Explanation:
The vapour pressure of a solution depends on the concentration of solute particles in the solution. In this case, the solute is a sparingly soluble salt AB3.
The solubility product (Ksp) of a sparingly soluble salt is a measure of its solubility in water. It is defined as the product of the concentrations of the ions in a saturated solution, each raised to the power of their stoichiometric coefficient.
The solubility product expression for AB3 can be written as:
AB3(s) ⇌ A3+(aq) + 3B-(aq)
The solubility product constant (Ksp) for AB3 can be given as:
Ksp = [A3+][B-]^3
Since AB3 is sparingly soluble, the concentration of AB3(s) in the saturated solution can be assumed to be constant. Therefore, the concentration of A3+ and B- ions in the saturated solution can be considered to be proportional to the vapour pressure of the solution.
Calculating the solubility product:
1. The vapour pressure of the saturated solution (P) is 17.25 mm.
2. The vapour pressure of the solvent (P°) is 17.33 mm.
3. The decrease in vapour pressure due to the presence of solute is given by:
ΔP = P° - P
= 17.33 - 17.25
= 0.08 mm
4. According to Raoult's law, the decrease in vapour pressure is directly proportional to the mole fraction of the solute particles in the solution.
ΔP ∝ x solute
5. Let the mole fraction of AB3(s) in the saturated solution be x. Then, the mole fraction of A3+ and B- ions can be assumed to be 3x and 3x, respectively.
ΔP ∝ x + 3x + 3x
ΔP ∝ 7x
6. Therefore, 7x = 0.08
x = 0.08/7
x = 0.0114
7. The concentration of A3+ and B- ions in the saturated solution can be assumed to be proportional to the mole fraction x.
[A3+] = 3x = 3(0.0114) = 0.0342
[B-] = 3x = 3(0.0114) = 0.0342
8. The solubility product (Ksp) can be calculated as the product of the concentrations of A3+ and B- ions.
Ksp = [A3+][B-]^3
= (0.0342)(0.0342)^3
= 0.0342 x 0.0342^3
= 0.0342 x 0.0000437
= 1.49 x 10^-6
Conclusion:
The solubility product (Ksp)
- Vapour pressure of the solvent (P°) = 17.33 mm
- Vapour pressure of the saturated solution (P) = 17.33 - 0.08 = 17.25 mm
Explanation:
The vapour pressure of a solution depends on the concentration of solute particles in the solution. In this case, the solute is a sparingly soluble salt AB3.
The solubility product (Ksp) of a sparingly soluble salt is a measure of its solubility in water. It is defined as the product of the concentrations of the ions in a saturated solution, each raised to the power of their stoichiometric coefficient.
The solubility product expression for AB3 can be written as:
AB3(s) ⇌ A3+(aq) + 3B-(aq)
The solubility product constant (Ksp) for AB3 can be given as:
Ksp = [A3+][B-]^3
Since AB3 is sparingly soluble, the concentration of AB3(s) in the saturated solution can be assumed to be constant. Therefore, the concentration of A3+ and B- ions in the saturated solution can be considered to be proportional to the vapour pressure of the solution.
Calculating the solubility product:
1. The vapour pressure of the saturated solution (P) is 17.25 mm.
2. The vapour pressure of the solvent (P°) is 17.33 mm.
3. The decrease in vapour pressure due to the presence of solute is given by:
ΔP = P° - P
= 17.33 - 17.25
= 0.08 mm
4. According to Raoult's law, the decrease in vapour pressure is directly proportional to the mole fraction of the solute particles in the solution.
ΔP ∝ x solute
5. Let the mole fraction of AB3(s) in the saturated solution be x. Then, the mole fraction of A3+ and B- ions can be assumed to be 3x and 3x, respectively.
ΔP ∝ x + 3x + 3x
ΔP ∝ 7x
6. Therefore, 7x = 0.08
x = 0.08/7
x = 0.0114
7. The concentration of A3+ and B- ions in the saturated solution can be assumed to be proportional to the mole fraction x.
[A3+] = 3x = 3(0.0114) = 0.0342
[B-] = 3x = 3(0.0114) = 0.0342
8. The solubility product (Ksp) can be calculated as the product of the concentrations of A3+ and B- ions.
Ksp = [A3+][B-]^3
= (0.0342)(0.0342)^3
= 0.0342 x 0.0342^3
= 0.0342 x 0.0000437
= 1.49 x 10^-6
Conclusion:
The solubility product (Ksp)
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A saturated aqueous solution of sparingly soluble salt AB3has the vapour pressure 0.08 mm lesser than the vapour pressure 17.33 mm of solvent at 25oC. The solubility product of AB3is :a)1.087 x 10-2b)1.48 x 10-4c)5.35 x 10-5d)4.56 x 10-4Correct answer is option 'D'. Can you explain this answer?
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A saturated aqueous solution of sparingly soluble salt AB3has the vapour pressure 0.08 mm lesser than the vapour pressure 17.33 mm of solvent at 25oC. The solubility product of AB3is :a)1.087 x 10-2b)1.48 x 10-4c)5.35 x 10-5d)4.56 x 10-4Correct answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A saturated aqueous solution of sparingly soluble salt AB3has the vapour pressure 0.08 mm lesser than the vapour pressure 17.33 mm of solvent at 25oC. The solubility product of AB3is :a)1.087 x 10-2b)1.48 x 10-4c)5.35 x 10-5d)4.56 x 10-4Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A saturated aqueous solution of sparingly soluble salt AB3has the vapour pressure 0.08 mm lesser than the vapour pressure 17.33 mm of solvent at 25oC. The solubility product of AB3is :a)1.087 x 10-2b)1.48 x 10-4c)5.35 x 10-5d)4.56 x 10-4Correct answer is option 'D'. Can you explain this answer?.
A saturated aqueous solution of sparingly soluble salt AB3has the vapour pressure 0.08 mm lesser than the vapour pressure 17.33 mm of solvent at 25oC. The solubility product of AB3is :a)1.087 x 10-2b)1.48 x 10-4c)5.35 x 10-5d)4.56 x 10-4Correct answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A saturated aqueous solution of sparingly soluble salt AB3has the vapour pressure 0.08 mm lesser than the vapour pressure 17.33 mm of solvent at 25oC. The solubility product of AB3is :a)1.087 x 10-2b)1.48 x 10-4c)5.35 x 10-5d)4.56 x 10-4Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A saturated aqueous solution of sparingly soluble salt AB3has the vapour pressure 0.08 mm lesser than the vapour pressure 17.33 mm of solvent at 25oC. The solubility product of AB3is :a)1.087 x 10-2b)1.48 x 10-4c)5.35 x 10-5d)4.56 x 10-4Correct answer is option 'D'. Can you explain this answer?.
Solutions for A saturated aqueous solution of sparingly soluble salt AB3has the vapour pressure 0.08 mm lesser than the vapour pressure 17.33 mm of solvent at 25oC. The solubility product of AB3is :a)1.087 x 10-2b)1.48 x 10-4c)5.35 x 10-5d)4.56 x 10-4Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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