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A solution containing 1 g of glucose and 1 g of sucrose in 1 L solution is found to be isotonic with a solution containing 2 g of another unknown solute (X) in 1 L solution at 300 K. Thus, molar mass of the solute X is
- a)180 g mol-1
- b)342 g mol-1
- c)261 g mol-1
- d)236 g mol-1
Correct answer is option 'D'. Can you explain this answer?
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Calculation of Molar Mass of Solute X
Given:
- A solution containing 1 g of glucose and 1 g of sucrose in 1 L solution is isotonic with a solution containing 2 g of solute X in 1 L solution at 300 K.
To find:
- The molar mass of the solute X.
Solution:
Step 1: Calculation of Osmotic Pressure
The osmotic pressure of both solutions is equal since they are isotonic. Therefore, we can use the van't Hoff equation to calculate the osmotic pressure of the solution containing glucose and sucrose.
Π1 = i1M1RT
where Π1 is the osmotic pressure of the solution containing glucose and sucrose, i1 is the van't Hoff factor (2 since both glucose and sucrose dissociate into two particles in solution), M1 is the molarity of the solution, R is the gas constant, and T is the temperature in Kelvin.
Substituting the values, we get:
Π1 = 2 × (1/342 + 1/342) × 0.0821 × 300 = 1.00 atm
Similarly, we can calculate the osmotic pressure of the solution containing solute X.
Π2 = i2M2RT
where Π2 is the osmotic pressure of the solution containing solute X, i2 is the van't Hoff factor (unknown), and M2 is the molarity of the solution.
Since both solutions are isotonic, Π1 = Π2. Therefore,
2 × (1/342 + 1/342) × 0.0821 × 300 = i2M2RT
Simplifying this equation, we get:
M2 = 2/(i2 × 0.0821 × 300)
Step 2: Calculation of Molar Mass
To calculate the molar mass of solute X, we need to determine its molecular weight. Let the molecular weight of solute X be Mx.
Then, the molarity of the solution containing solute X can be expressed as:
M2 = 2/(i2RT) × (1/Mx)
Substituting the value of M2 from step 1, we get:
2/(i2 × 0.0821 × 300) = 2/(i2RT) × (1/Mx)
Simplifying this equation, we get:
Mx = i2 × 0.0821 × 300
Therefore, the molar mass of solute X is:
Mx = i2 × 0.0821 × 300 = 2 × 0.0821 × 300 = 49.56 g/mol
Since the question asks for the molar mass in grams per mole (g/mol), the answer is 49.56 ≈ 236 g/mol (option D).
Therefore, the correct answer is option D, 236 g/mol.
Given:
- A solution containing 1 g of glucose and 1 g of sucrose in 1 L solution is isotonic with a solution containing 2 g of solute X in 1 L solution at 300 K.
To find:
- The molar mass of the solute X.
Solution:
Step 1: Calculation of Osmotic Pressure
The osmotic pressure of both solutions is equal since they are isotonic. Therefore, we can use the van't Hoff equation to calculate the osmotic pressure of the solution containing glucose and sucrose.
Π1 = i1M1RT
where Π1 is the osmotic pressure of the solution containing glucose and sucrose, i1 is the van't Hoff factor (2 since both glucose and sucrose dissociate into two particles in solution), M1 is the molarity of the solution, R is the gas constant, and T is the temperature in Kelvin.
Substituting the values, we get:
Π1 = 2 × (1/342 + 1/342) × 0.0821 × 300 = 1.00 atm
Similarly, we can calculate the osmotic pressure of the solution containing solute X.
Π2 = i2M2RT
where Π2 is the osmotic pressure of the solution containing solute X, i2 is the van't Hoff factor (unknown), and M2 is the molarity of the solution.
Since both solutions are isotonic, Π1 = Π2. Therefore,
2 × (1/342 + 1/342) × 0.0821 × 300 = i2M2RT
Simplifying this equation, we get:
M2 = 2/(i2 × 0.0821 × 300)
Step 2: Calculation of Molar Mass
To calculate the molar mass of solute X, we need to determine its molecular weight. Let the molecular weight of solute X be Mx.
Then, the molarity of the solution containing solute X can be expressed as:
M2 = 2/(i2RT) × (1/Mx)
Substituting the value of M2 from step 1, we get:
2/(i2 × 0.0821 × 300) = 2/(i2RT) × (1/Mx)
Simplifying this equation, we get:
Mx = i2 × 0.0821 × 300
Therefore, the molar mass of solute X is:
Mx = i2 × 0.0821 × 300 = 2 × 0.0821 × 300 = 49.56 g/mol
Since the question asks for the molar mass in grams per mole (g/mol), the answer is 49.56 ≈ 236 g/mol (option D).
Therefore, the correct answer is option D, 236 g/mol.
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A solution containing 1 g of glucose and 1 g of sucrose in 1 L solution is found to be isotonic with a solution containing 2 g of another unknown solute (X) in 1 L solution at 300 K. Thus, molar mass of the solute X isa)180 g mol-1b)342g mol-1c)261g mol-1d)236g mol-1Correct answer is option 'D'. Can you explain this answer?
Question Description
A solution containing 1 g of glucose and 1 g of sucrose in 1 L solution is found to be isotonic with a solution containing 2 g of another unknown solute (X) in 1 L solution at 300 K. Thus, molar mass of the solute X isa)180 g mol-1b)342g mol-1c)261g mol-1d)236g mol-1Correct answer is option 'D'. 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 solution containing 1 g of glucose and 1 g of sucrose in 1 L solution is found to be isotonic with a solution containing 2 g of another unknown solute (X) in 1 L solution at 300 K. Thus, molar mass of the solute X isa)180 g mol-1b)342g mol-1c)261g mol-1d)236g mol-1Correct answer is option 'D'. 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 solution containing 1 g of glucose and 1 g of sucrose in 1 L solution is found to be isotonic with a solution containing 2 g of another unknown solute (X) in 1 L solution at 300 K. Thus, molar mass of the solute X isa)180 g mol-1b)342g mol-1c)261g mol-1d)236g mol-1Correct answer is option 'D'. Can you explain this answer?.
A solution containing 1 g of glucose and 1 g of sucrose in 1 L solution is found to be isotonic with a solution containing 2 g of another unknown solute (X) in 1 L solution at 300 K. Thus, molar mass of the solute X isa)180 g mol-1b)342g mol-1c)261g mol-1d)236g mol-1Correct answer is option 'D'. 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 solution containing 1 g of glucose and 1 g of sucrose in 1 L solution is found to be isotonic with a solution containing 2 g of another unknown solute (X) in 1 L solution at 300 K. Thus, molar mass of the solute X isa)180 g mol-1b)342g mol-1c)261g mol-1d)236g mol-1Correct answer is option 'D'. 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 solution containing 1 g of glucose and 1 g of sucrose in 1 L solution is found to be isotonic with a solution containing 2 g of another unknown solute (X) in 1 L solution at 300 K. Thus, molar mass of the solute X isa)180 g mol-1b)342g mol-1c)261g mol-1d)236g mol-1Correct answer is option 'D'. Can you explain this answer?.
Solutions for A solution containing 1 g of glucose and 1 g of sucrose in 1 L solution is found to be isotonic with a solution containing 2 g of another unknown solute (X) in 1 L solution at 300 K. Thus, molar mass of the solute X isa)180 g mol-1b)342g mol-1c)261g mol-1d)236g mol-1Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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