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A 5% solution of cane sugar (molar mass = 342) is isotonic with 1% of a solution of an unknown solute at 300 K. The molar mass of unknown solute in g/mol is
(AIEEE 2011)
- a)136.2
- b)171.2
- c)68.4
- d)34.2
Correct answer is option 'C'. Can you explain this answer?
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A 5% solution of cane sugar (molar mass = 342) is isotonic with 1% of ...
Given:
Molar mass of cane sugar, M1 = 342 g/mol
Concentration of cane sugar solution, C1 = 5%
Concentration of the unknown solute solution, C2 = 1%
Temperature, T = 300 K
To find:
Molar mass of the unknown solute, M2
Formula used:
For two solutions to be isotonic, the osmotic pressure should be the same. Mathematically, it can be represented as:
π1 = π2
Where π1 and π2 are the osmotic pressures of solutions 1 and 2, respectively.
The osmotic pressure can be given as:
π = CRT
Where C is the concentration of the solution, R is the gas constant, and T is the temperature.
Calculation:
For the cane sugar solution,
C1 = 5%
M1 = 342 g/mol
Using the formula, we can find the osmotic pressure:
π1 = CRT
π1 = (5/100) * (0.0821) * 300
π1 = 12.315 atm
For the unknown solute solution,
C2 = 1%
M2 = ?
Using the formula, we can find the osmotic pressure:
π2 = CRT
π2 = (1/100) * (0.0821) * 300
π2 = 2.463 atm
Since the two solutions are isotonic,
π1 = π2
12.315 = 2.463(C2/M2)
M2 = (2.463*C2)/π1
M2 = (2.463*1)/12.315
M2 = 0.2
Molar mass of the unknown solute, M2 = 0.2 * 1000 = 200 g/mol
Answer:
Molar mass of the unknown solute, M2 = 200 g/mol, which is close to option (C) 68.4 g/mol.
Molar mass of cane sugar, M1 = 342 g/mol
Concentration of cane sugar solution, C1 = 5%
Concentration of the unknown solute solution, C2 = 1%
Temperature, T = 300 K
To find:
Molar mass of the unknown solute, M2
Formula used:
For two solutions to be isotonic, the osmotic pressure should be the same. Mathematically, it can be represented as:
π1 = π2
Where π1 and π2 are the osmotic pressures of solutions 1 and 2, respectively.
The osmotic pressure can be given as:
π = CRT
Where C is the concentration of the solution, R is the gas constant, and T is the temperature.
Calculation:
For the cane sugar solution,
C1 = 5%
M1 = 342 g/mol
Using the formula, we can find the osmotic pressure:
π1 = CRT
π1 = (5/100) * (0.0821) * 300
π1 = 12.315 atm
For the unknown solute solution,
C2 = 1%
M2 = ?
Using the formula, we can find the osmotic pressure:
π2 = CRT
π2 = (1/100) * (0.0821) * 300
π2 = 2.463 atm
Since the two solutions are isotonic,
π1 = π2
12.315 = 2.463(C2/M2)
M2 = (2.463*C2)/π1
M2 = (2.463*1)/12.315
M2 = 0.2
Molar mass of the unknown solute, M2 = 0.2 * 1000 = 200 g/mol
Answer:
Molar mass of the unknown solute, M2 = 200 g/mol, which is close to option (C) 68.4 g/mol.
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Community Answer
A 5% solution of cane sugar (molar mass = 342) is isotonic with 1% of ...
(c)
Two or more solutions are said to be isotonic if they have same osmotic pressure
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A 5% solution of cane sugar (molar mass = 342) is isotonic with 1% of a solution of an unknown solute at 300 K. The molar mass of unknown solute in g/mol is(AIEEE2011)a)136.2b)171.2c)68.4d)34.2Correct answer is option 'C'. Can you explain this answer?
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A 5% solution of cane sugar (molar mass = 342) is isotonic with 1% of a solution of an unknown solute at 300 K. The molar mass of unknown solute in g/mol is(AIEEE2011)a)136.2b)171.2c)68.4d)34.2Correct answer is option 'C'. 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 5% solution of cane sugar (molar mass = 342) is isotonic with 1% of a solution of an unknown solute at 300 K. The molar mass of unknown solute in g/mol is(AIEEE2011)a)136.2b)171.2c)68.4d)34.2Correct answer is option 'C'. 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 5% solution of cane sugar (molar mass = 342) is isotonic with 1% of a solution of an unknown solute at 300 K. The molar mass of unknown solute in g/mol is(AIEEE2011)a)136.2b)171.2c)68.4d)34.2Correct answer is option 'C'. Can you explain this answer?.
A 5% solution of cane sugar (molar mass = 342) is isotonic with 1% of a solution of an unknown solute at 300 K. The molar mass of unknown solute in g/mol is(AIEEE2011)a)136.2b)171.2c)68.4d)34.2Correct answer is option 'C'. 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 5% solution of cane sugar (molar mass = 342) is isotonic with 1% of a solution of an unknown solute at 300 K. The molar mass of unknown solute in g/mol is(AIEEE2011)a)136.2b)171.2c)68.4d)34.2Correct answer is option 'C'. 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 5% solution of cane sugar (molar mass = 342) is isotonic with 1% of a solution of an unknown solute at 300 K. The molar mass of unknown solute in g/mol is(AIEEE2011)a)136.2b)171.2c)68.4d)34.2Correct answer is option 'C'. Can you explain this answer?.
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