Q57.The solubility product of a sparingly soluble salt A2X3 is 1. 1 × 10−23 . If specific conductance of the solution is 3 × 10−5 Sm−1 , the limiting molar conductivity of the solution is x × 10−3 S m2 mol−1 . The value of x is
What This Question Tests
This question requires applying Raoult's Law and Dalton's Law of partial pressures to calculate the mole fraction of a component in the vapor phase from its liquid phase mole fraction and pure vapor pressures.
Concepts Tested
Formulas Used
P_total = P_A^0 X_A + P_B^0 X_B
Y_B = P_B / P_total
📚 NCERT Sections This Tests
2.8 — The Conductivity Of 0.20 M Solution Of Kcl At 298 K Is 0.0248 S Cm–1. Calculate
Chemistry Class 11 · Chapter 2
2.8 The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm–1. Calculate its molar conductivity.
1.27 — If The Solubility Product Of Cus Is 6 × 10–16, Calculate The Maximum Molarity Of
Chemistry Class 11 · Chapter 1
1.27 If the solubility product of CuS is 6 × 10–16, calculate the maximum molarity of CuS in aqueous solution.
2.7 — Define Conductivity And Molar Conductivity For The Solution Of An Electrolyte.
Chemistry Class 11 · Chapter 2
2.7 Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
📋 Question Details
- Chapter
- Solutions
- Topic
- Raoult's Law
- Year
- 2022
- Shift
- 28 Jun Shift 1
- Q Number
- Q57
- Type
- Numerical
- NCERT Ref
- Class 12 Chemistry Ch 2: Solutions
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