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At 298 K, the vapour pressure of pure water is 23.76 mm Hg and that of an aqueous solution is 22.98 mm Hg. Thus, molality of solution is (assume dilute solution)
- a)0.033
- b)0.67
- c)1.83
- d)5.55
Correct answer is option 'C'. Can you explain this answer?
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At 298 K, the vapour pressure of pure water is 23.76 mm Hg and that of...
(c)
By Raoult’s law for a non-volatile solute in volatile solvent.
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At 298 K, the vapour pressure of pure water is 23.76 mm Hg and that of...
Given:
Vapour pressure of pure water (P°) = 23.76 mm Hg
Vapour pressure of aqueous solution (P) = 22.98 mm Hg
To find:
Molality of the solution
Formula used:
Raoult's law: P = P° x mole fraction of solvent
Solution:
1. To find mole fraction of solvent:
Using Raoult's law, we have:
P = P° x mole fraction of solvent
Mole fraction of solvent = P/P°
Mole fraction of solvent = 22.98/23.76
Mole fraction of solvent = 0.9678
2. To find molality of the solution:
Molality (m) = moles of solute / mass of solvent in kg
Since the solution is dilute, we can assume that the mass of the solvent is equal to the mass of the solution.
Let's assume that we have 1000g (or 1kg) of the solution.
Mass of solvent = 1000g - mass of solute
We need to find moles of solute to calculate molality.
Moles of solute = (mass of solute / molar mass of solute)
Since the solute is not given, we can assume it to be a non-volatile solute like glucose.
Molar mass of glucose (C6H12O6) = (6 x 12.01) + (12 x 1.01) + (6 x 16.00) = 180.18 g/mol
Let's assume that we have 'x' grams of glucose.
Moles of glucose = (x/180.18)
Using the formula for molality, we have:
m = (x/180.18) / 0.9678
m = (x/180.18) / (22.98/23.76)
m = (x/180.18) x (23.76/22.98)
m = (x/180.18) x 1.034
m = x/174.98
3. To find the value of 'x':
We know that the mass of solvent = 1000g - mass of solute = (1000 - x)g
Using the formula for mole fraction of solute, we have:
Mole fraction of solute = (moles of solute) / [(moles of solute) + (moles of solvent)]
Mole fraction of solute = (x/180.18) / [(x/180.18) + (1000 - x)/18.02]
Mole fraction of solute = (x/180.18) / [(1000 + 180.18 - x)/180.18]
Mole fraction of solute = (x/180.18) x (180.18 / (1000 + 180.18 - x))
Mole fraction of solute = x / (1000 + 180.18 - x)
Using the formula for molality, we have:
m = (x/180.18) / (1 - x/(1000 + 180.18))
Solving for 'x',
Vapour pressure of pure water (P°) = 23.76 mm Hg
Vapour pressure of aqueous solution (P) = 22.98 mm Hg
To find:
Molality of the solution
Formula used:
Raoult's law: P = P° x mole fraction of solvent
Solution:
1. To find mole fraction of solvent:
Using Raoult's law, we have:
P = P° x mole fraction of solvent
Mole fraction of solvent = P/P°
Mole fraction of solvent = 22.98/23.76
Mole fraction of solvent = 0.9678
2. To find molality of the solution:
Molality (m) = moles of solute / mass of solvent in kg
Since the solution is dilute, we can assume that the mass of the solvent is equal to the mass of the solution.
Let's assume that we have 1000g (or 1kg) of the solution.
Mass of solvent = 1000g - mass of solute
We need to find moles of solute to calculate molality.
Moles of solute = (mass of solute / molar mass of solute)
Since the solute is not given, we can assume it to be a non-volatile solute like glucose.
Molar mass of glucose (C6H12O6) = (6 x 12.01) + (12 x 1.01) + (6 x 16.00) = 180.18 g/mol
Let's assume that we have 'x' grams of glucose.
Moles of glucose = (x/180.18)
Using the formula for molality, we have:
m = (x/180.18) / 0.9678
m = (x/180.18) / (22.98/23.76)
m = (x/180.18) x (23.76/22.98)
m = (x/180.18) x 1.034
m = x/174.98
3. To find the value of 'x':
We know that the mass of solvent = 1000g - mass of solute = (1000 - x)g
Using the formula for mole fraction of solute, we have:
Mole fraction of solute = (moles of solute) / [(moles of solute) + (moles of solvent)]
Mole fraction of solute = (x/180.18) / [(x/180.18) + (1000 - x)/18.02]
Mole fraction of solute = (x/180.18) / [(1000 + 180.18 - x)/180.18]
Mole fraction of solute = (x/180.18) x (180.18 / (1000 + 180.18 - x))
Mole fraction of solute = x / (1000 + 180.18 - x)
Using the formula for molality, we have:
m = (x/180.18) / (1 - x/(1000 + 180.18))
Solving for 'x',
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At 298 K, the vapour pressure of pure water is 23.76 mm Hg and that of an aqueous solution is 22.98 mm Hg. Thus, molality of solution is (assume dilute solution)a)0.033b)0.67c)1.83d)5.55Correct answer is option 'C'. Can you explain this answer?
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At 298 K, the vapour pressure of pure water is 23.76 mm Hg and that of an aqueous solution is 22.98 mm Hg. Thus, molality of solution is (assume dilute solution)a)0.033b)0.67c)1.83d)5.55Correct answer is option 'C'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about At 298 K, the vapour pressure of pure water is 23.76 mm Hg and that of an aqueous solution is 22.98 mm Hg. Thus, molality of solution is (assume dilute solution)a)0.033b)0.67c)1.83d)5.55Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 298 K, the vapour pressure of pure water is 23.76 mm Hg and that of an aqueous solution is 22.98 mm Hg. Thus, molality of solution is (assume dilute solution)a)0.033b)0.67c)1.83d)5.55Correct answer is option 'C'. Can you explain this answer?.
At 298 K, the vapour pressure of pure water is 23.76 mm Hg and that of an aqueous solution is 22.98 mm Hg. Thus, molality of solution is (assume dilute solution)a)0.033b)0.67c)1.83d)5.55Correct answer is option 'C'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about At 298 K, the vapour pressure of pure water is 23.76 mm Hg and that of an aqueous solution is 22.98 mm Hg. Thus, molality of solution is (assume dilute solution)a)0.033b)0.67c)1.83d)5.55Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 298 K, the vapour pressure of pure water is 23.76 mm Hg and that of an aqueous solution is 22.98 mm Hg. Thus, molality of solution is (assume dilute solution)a)0.033b)0.67c)1.83d)5.55Correct answer is option 'C'. Can you explain this answer?.
Solutions for At 298 K, the vapour pressure of pure water is 23.76 mm Hg and that of an aqueous solution is 22.98 mm Hg. Thus, molality of solution is (assume dilute solution)a)0.033b)0.67c)1.83d)5.55Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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At 298 K, the vapour pressure of pure water is 23.76 mm Hg and that of an aqueous solution is 22.98 mm Hg. Thus, molality of solution is (assume dilute solution)a)0.033b)0.67c)1.83d)5.55Correct answer is option 'C'. Can you explain this answer?, a detailed solution for At 298 K, the vapour pressure of pure water is 23.76 mm Hg and that of an aqueous solution is 22.98 mm Hg. Thus, molality of solution is (assume dilute solution)a)0.033b)0.67c)1.83d)5.55Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of At 298 K, the vapour pressure of pure water is 23.76 mm Hg and that of an aqueous solution is 22.98 mm Hg. Thus, molality of solution is (assume dilute solution)a)0.033b)0.67c)1.83d)5.55Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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