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A 5% solution of cane sugar (molar mass = 342 g mol-1) is isotonic with 1 % of a solution of an unknown non-electrolyte solute at 300 K. The molar mass of unknown solute is (in g mol-1)
[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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Isotonic Solutions and Colligative Properties:
An isotonic solution is a solution that has the same osmotic pressure as another solution. In other words, it is a solution in which the concentration of solute particles is the same as that of another solution. The osmotic pressure of a solution depends on the number of solute particles, rather than their chemical nature. This property is known as a colligative property.
The formula to calculate the osmotic pressure of a solution is:
π = (n/V)RT
Where:
π is the osmotic pressure
n/V is the concentration of solute particles in moles per liter (mol/L)
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature in Kelvin (K)
Isotonicity between Cane Sugar and Unknown Solute:
Given that the 5% cane sugar solution is isotonic with a 1% solution of the unknown solute, we can assume that the osmotic pressure of both solutions is the same. Therefore, we can equate the osmotic pressure equations for both solutions:
For the cane sugar solution:
π1 = (n1/V1)RT
For the unknown solute solution:
π2 = (n2/V2)RT
Since the osmotic pressures are the same, we can write:
(n1/V1)RT = (n2/V2)RT
Simplifying, we get:
(n1/V1) = (n2/V2)
Since the molar mass of cane sugar (M1) is given as 342 g/mol, and the concentration of the cane sugar solution is 5%, we can write:
(n1/V1) = (0.05/M1)
Similarly, for the unknown solute, let the molar mass be M2. The concentration of the unknown solute solution is given as 1%, so we can write:
(n2/V2) = (0.01/M2)
Equating these expressions, we have:
(0.05/M1) = (0.01/M2)
Solving for M2, we get:
M2 = (0.01/M1) * (M2/0.05)
M2 = (0.01/0.05) * M1
M2 = 0.2 * M1
Substituting the molar mass of cane sugar (M1 = 342 g/mol), we get:
M2 = 0.2 * 342
M2 = 68.4 g/mol
Therefore, the molar mass of the unknown solute is 68.4 g/mol, which corresponds to option 'C'.
An isotonic solution is a solution that has the same osmotic pressure as another solution. In other words, it is a solution in which the concentration of solute particles is the same as that of another solution. The osmotic pressure of a solution depends on the number of solute particles, rather than their chemical nature. This property is known as a colligative property.
The formula to calculate the osmotic pressure of a solution is:
π = (n/V)RT
Where:
π is the osmotic pressure
n/V is the concentration of solute particles in moles per liter (mol/L)
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature in Kelvin (K)
Isotonicity between Cane Sugar and Unknown Solute:
Given that the 5% cane sugar solution is isotonic with a 1% solution of the unknown solute, we can assume that the osmotic pressure of both solutions is the same. Therefore, we can equate the osmotic pressure equations for both solutions:
For the cane sugar solution:
π1 = (n1/V1)RT
For the unknown solute solution:
π2 = (n2/V2)RT
Since the osmotic pressures are the same, we can write:
(n1/V1)RT = (n2/V2)RT
Simplifying, we get:
(n1/V1) = (n2/V2)
Since the molar mass of cane sugar (M1) is given as 342 g/mol, and the concentration of the cane sugar solution is 5%, we can write:
(n1/V1) = (0.05/M1)
Similarly, for the unknown solute, let the molar mass be M2. The concentration of the unknown solute solution is given as 1%, so we can write:
(n2/V2) = (0.01/M2)
Equating these expressions, we have:
(0.05/M1) = (0.01/M2)
Solving for M2, we get:
M2 = (0.01/M1) * (M2/0.05)
M2 = (0.01/0.05) * M1
M2 = 0.2 * M1
Substituting the molar mass of cane sugar (M1 = 342 g/mol), we get:
M2 = 0.2 * 342
M2 = 68.4 g/mol
Therefore, the molar mass of the unknown solute is 68.4 g/mol, which corresponds to option 'C'.
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A 5% solution of cane sugar (molar mass = 342 g mol-1) is isotonic with 1 % of a solution of an unknown non-electrolyte solute at 300 K. The molar mass of unknown solute is (in g mol-1)[Aieee 2011]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 g mol-1) is isotonic with 1 % of a solution of an unknown non-electrolyte solute at 300 K. The molar mass of unknown solute is (in g mol-1)[Aieee 2011]a)136.2b)171.2c)68.4d)34.2Correct 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 A 5% solution of cane sugar (molar mass = 342 g mol-1) is isotonic with 1 % of a solution of an unknown non-electrolyte solute at 300 K. The molar mass of unknown solute is (in g mol-1)[Aieee 2011]a)136.2b)171.2c)68.4d)34.2Correct 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 A 5% solution of cane sugar (molar mass = 342 g mol-1) is isotonic with 1 % of a solution of an unknown non-electrolyte solute at 300 K. The molar mass of unknown solute is (in g mol-1)[Aieee 2011]a)136.2b)171.2c)68.4d)34.2Correct answer is option 'C'. Can you explain this answer?.
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