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Passage l
W ater boils at 100°C at 1 atm pressure. Latent heat of vaporisation (ΔH ) = 10 kcal mol-1.
Q.
Lowering of vapour pressure of 0.2 molal urea solution at 40°C is
- a)0.2087 torr
- b)0.2912 torr
- c)1.2000 torr
- d)0.6920 torr
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Passage lW ater boils at 100C at 1 atm pressure. Latent heat of vapori...
Variation of vapour pressure with temperature is given Clapeyron-Clausius equation.
p1 = 0.07656 atm
p1 = 0.07656 x 760 torr = 58.19torr
Vapour pressure at 40°C(313 K) = 58.19 torr
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Passage lW ater boils at 100C at 1 atm pressure. Latent heat of vapori...
Explanation:
Boiling Point Elevation:
- Boiling point elevation is given by the formula: ΔTb = i * Kf * m, where i is the Van't Hoff factor, Kf is the ebullioscopic constant, and m is the molality of the solution.
- Given that the latent heat of vaporisation (Hv) = 10 kcal mol-1, we can calculate the boiling point elevation using the formula: ΔTb = (Hv / R) * ln(X2/X1), where R is the gas constant, X1 is the initial mole fraction, and X2 is the final mole fraction.
Calculation:
- Since the urea solution is 0.2 molal, the molality (m) = 0.2 mol/kg.
- At 40°C, the vapour pressure of pure water is 55.3 torr. The lowering of vapour pressure due to the urea solution can be calculated using the formula: ΔP = i * m * Kf * Tb, where Tb is the boiling point of the solution.
- With the given values, we can calculate the lowering of vapour pressure to be ΔP = 0.2087 torr.
Therefore, the correct answer is option 'A' (0.2087 torr).
Boiling Point Elevation:
- Boiling point elevation is given by the formula: ΔTb = i * Kf * m, where i is the Van't Hoff factor, Kf is the ebullioscopic constant, and m is the molality of the solution.
- Given that the latent heat of vaporisation (Hv) = 10 kcal mol-1, we can calculate the boiling point elevation using the formula: ΔTb = (Hv / R) * ln(X2/X1), where R is the gas constant, X1 is the initial mole fraction, and X2 is the final mole fraction.
Calculation:
- Since the urea solution is 0.2 molal, the molality (m) = 0.2 mol/kg.
- At 40°C, the vapour pressure of pure water is 55.3 torr. The lowering of vapour pressure due to the urea solution can be calculated using the formula: ΔP = i * m * Kf * Tb, where Tb is the boiling point of the solution.
- With the given values, we can calculate the lowering of vapour pressure to be ΔP = 0.2087 torr.
Therefore, the correct answer is option 'A' (0.2087 torr).
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Passage lW ater boils at 100C at 1 atm pressure. Latent heat of vaporisation (H ) = 10 kcal mol-1.Q.Lowering of vapour pressure of 0.2 molal urea solution at 40C isa)0.2087 torrb)0.2912 torrc)1.2000 torrd)0.6920 torrCorrect answer is option 'A'. Can you explain this answer?
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Passage lW ater boils at 100C at 1 atm pressure. Latent heat of vaporisation (H ) = 10 kcal mol-1.Q.Lowering of vapour pressure of 0.2 molal urea solution at 40C isa)0.2087 torrb)0.2912 torrc)1.2000 torrd)0.6920 torrCorrect answer is option 'A'. 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 Passage lW ater boils at 100C at 1 atm pressure. Latent heat of vaporisation (H ) = 10 kcal mol-1.Q.Lowering of vapour pressure of 0.2 molal urea solution at 40C isa)0.2087 torrb)0.2912 torrc)1.2000 torrd)0.6920 torrCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Passage lW ater boils at 100C at 1 atm pressure. Latent heat of vaporisation (H ) = 10 kcal mol-1.Q.Lowering of vapour pressure of 0.2 molal urea solution at 40C isa)0.2087 torrb)0.2912 torrc)1.2000 torrd)0.6920 torrCorrect answer is option 'A'. Can you explain this answer?.
Passage lW ater boils at 100C at 1 atm pressure. Latent heat of vaporisation (H ) = 10 kcal mol-1.Q.Lowering of vapour pressure of 0.2 molal urea solution at 40C isa)0.2087 torrb)0.2912 torrc)1.2000 torrd)0.6920 torrCorrect answer is option 'A'. 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 Passage lW ater boils at 100C at 1 atm pressure. Latent heat of vaporisation (H ) = 10 kcal mol-1.Q.Lowering of vapour pressure of 0.2 molal urea solution at 40C isa)0.2087 torrb)0.2912 torrc)1.2000 torrd)0.6920 torrCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Passage lW ater boils at 100C at 1 atm pressure. Latent heat of vaporisation (H ) = 10 kcal mol-1.Q.Lowering of vapour pressure of 0.2 molal urea solution at 40C isa)0.2087 torrb)0.2912 torrc)1.2000 torrd)0.6920 torrCorrect answer is option 'A'. Can you explain this answer?.
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