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A 0.2 molal aqueous solution of a weak acid (HX) is 20 percent ionised. The freezing point of this solution is (Given kf = 1.86º C kg mol–1 for water) :
- a)–0.45ºC
- b)– 0.90ºC
- c)–0.31 ºC
- d)– 0.53ºC
Correct answer is option 'A'. Can you explain this answer?
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A 0.2 molal aqueous solution of a weak acid (HX) is 20 percent ionised...
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A 0.2 molal aqueous solution of a weak acid (HX) is 20 percent ionised...
Given Data
- Molality of the solution (m) = 0.2 mol/kg
- Percent ionization = 20%
- Freezing point depression constant (kf) = 1.86 °C kg/mol
Ionization Calculation
- Weak Acid Ionization:
- The weak acid (HX) ionizes to give H+ and X-.
- For 0.2 molal solution, 20% ionization means:
- Moles of HX that ionize = 0.2 mol × 20% = 0.04 mol
- Total Moles in Solution:
- Remaining non-ionized HX = 0.2 mol - 0.04 mol = 0.16 mol
- Total moles = 0.04 mol (H+) + 0.04 mol (X-) + 0.16 mol (undissociated HX) = 0.24 mol
Freezing Point Depression Calculation
- Using the formula for freezing point depression:
- ΔTf = i × kf × m
- Where i = van 't Hoff factor, which is the total number of particles in solution.
- Here, i = 1 (for undissociated HX) + 0.04 + 0.04 = 1.24
- Calculation of ΔTf:
- ΔTf = 1.24 × 1.86 °C kg/mol × 0.2 mol/kg
- ΔTf = 1.24 × 1.86 × 0.2 = 0.46 °C (approximately)
Final Freezing Point
- Freezing Point of Water: 0 °C
- Final Freezing Point:
- Freezing point = 0 °C - ΔTf = 0 °C - 0.46 °C = -0.46 °C (approximately -0.45 °C).
Thus, the freezing point of the solution is approximately -0.45 °C, making option 'A' the correct answer.
- Molality of the solution (m) = 0.2 mol/kg
- Percent ionization = 20%
- Freezing point depression constant (kf) = 1.86 °C kg/mol
Ionization Calculation
- Weak Acid Ionization:
- The weak acid (HX) ionizes to give H+ and X-.
- For 0.2 molal solution, 20% ionization means:
- Moles of HX that ionize = 0.2 mol × 20% = 0.04 mol
- Total Moles in Solution:
- Remaining non-ionized HX = 0.2 mol - 0.04 mol = 0.16 mol
- Total moles = 0.04 mol (H+) + 0.04 mol (X-) + 0.16 mol (undissociated HX) = 0.24 mol
Freezing Point Depression Calculation
- Using the formula for freezing point depression:
- ΔTf = i × kf × m
- Where i = van 't Hoff factor, which is the total number of particles in solution.
- Here, i = 1 (for undissociated HX) + 0.04 + 0.04 = 1.24
- Calculation of ΔTf:
- ΔTf = 1.24 × 1.86 °C kg/mol × 0.2 mol/kg
- ΔTf = 1.24 × 1.86 × 0.2 = 0.46 °C (approximately)
Final Freezing Point
- Freezing Point of Water: 0 °C
- Final Freezing Point:
- Freezing point = 0 °C - ΔTf = 0 °C - 0.46 °C = -0.46 °C (approximately -0.45 °C).
Thus, the freezing point of the solution is approximately -0.45 °C, making option 'A' the correct answer.
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A 0.2 molal aqueous solution of a weak acid (HX) is 20 percent ionised. The freezing point of this solution is (Given kf = 1.86º C kg mol–1 for water) :a) –0.45ºCb) – 0.90ºCc) –0.31 ºCd) – 0.53ºCCorrect answer is option 'A'. Can you explain this answer?
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A 0.2 molal aqueous solution of a weak acid (HX) is 20 percent ionised. The freezing point of this solution is (Given kf = 1.86º C kg mol–1 for water) :a) –0.45ºCb) – 0.90ºCc) –0.31 ºCd) – 0.53ºCCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A 0.2 molal aqueous solution of a weak acid (HX) is 20 percent ionised. The freezing point of this solution is (Given kf = 1.86º C kg mol–1 for water) :a) –0.45ºCb) – 0.90ºCc) –0.31 ºCd) – 0.53ºCCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 0.2 molal aqueous solution of a weak acid (HX) is 20 percent ionised. The freezing point of this solution is (Given kf = 1.86º C kg mol–1 for water) :a) –0.45ºCb) – 0.90ºCc) –0.31 ºCd) – 0.53ºCCorrect answer is option 'A'. Can you explain this answer?.
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