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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?
Most Upvoted Answer
A saturated aqueous solution of sparingly soluble salt AB3has the vapo...
Understanding the Problem
To determine the solubility product (Ksp) of the sparingly soluble salt AB3, we need to analyze the given information about the saturated solution and its effect on vapor pressure.
Given Data
- Vapor pressure of solvent (P0) = 17.33 mm
- Decrease in vapor pressure (ΔP) = 0.08 mm
- Vapor pressure of the solution (P) = P0 - ΔP = 17.33 - 0.08 = 17.25 mm
Calculating Mole Fraction of Solvent
Using Raoult's law, we find:
- ΔP = Xsolute * P0
Where:
- Xsolute = mole fraction of the solute (AB3)
Thus,
Xsolute = ΔP / P0 = 0.08 / 17.33 = 0.0046
Finding Mole Fraction of Solvent
- Xsolvent = 1 - Xsolute
- Xsolvent = 1 - 0.0046 = 0.9954
Determining the Concentration of AB3
Assuming the salt AB3 dissociates in solution according to the equation:
AB3 ⇌ A + 3B
Let the solubility of AB3 be 's'. The concentration of ions produced is:
- [A] = s
- [B] = 3s
Then, Ksp can be expressed as:
Ksp = [A] * [B]^3 = s * (3s)^3 = 27s^4
Relating Mole Fraction to Solubility
Using the relationship between the mole fraction of the solute and the solubility:
Xsolute = s / (s + 1) (assuming the solvent's molar mass is considerably larger)
Since Xsolute is small, we can approximate:
s ≈ Xsolute
Substituting the value:
s ≈ 0.0046
Calculating Ksp
Now, substituting s back into the Ksp expression:
Ksp = 27 * (0.0046)^4 = 4.56 x 10^-4
Thus, the correct answer is option 'D'.
To determine the solubility product (Ksp) of the sparingly soluble salt AB3, we need to analyze the given information about the saturated solution and its effect on vapor pressure.
Given Data
- Vapor pressure of solvent (P0) = 17.33 mm
- Decrease in vapor pressure (ΔP) = 0.08 mm
- Vapor pressure of the solution (P) = P0 - ΔP = 17.33 - 0.08 = 17.25 mm
Calculating Mole Fraction of Solvent
Using Raoult's law, we find:
- ΔP = Xsolute * P0
Where:
- Xsolute = mole fraction of the solute (AB3)
Thus,
Xsolute = ΔP / P0 = 0.08 / 17.33 = 0.0046
Finding Mole Fraction of Solvent
- Xsolvent = 1 - Xsolute
- Xsolvent = 1 - 0.0046 = 0.9954
Determining the Concentration of AB3
Assuming the salt AB3 dissociates in solution according to the equation:
AB3 ⇌ A + 3B
Let the solubility of AB3 be 's'. The concentration of ions produced is:
- [A] = s
- [B] = 3s
Then, Ksp can be expressed as:
Ksp = [A] * [B]^3 = s * (3s)^3 = 27s^4
Relating Mole Fraction to Solubility
Using the relationship between the mole fraction of the solute and the solubility:
Xsolute = s / (s + 1) (assuming the solvent's molar mass is considerably larger)
Since Xsolute is small, we can approximate:
s ≈ Xsolute
Substituting the value:
s ≈ 0.0046
Calculating Ksp
Now, substituting s back into the Ksp expression:
Ksp = 27 * (0.0046)^4 = 4.56 x 10^-4
Thus, the correct answer is option 'D'.
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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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