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5.25% solution of a substance is isotonic with a 21 .5% solution of urea (NH2CONH2 )by weight of solution in the same solvent at 300 K. If the densities of both the solutions are assumed to be equal to 1.0 g cm-3 then, molar mass of the substance will be
- a)90g mol-1
- b)115g mol-1
- c)105g mol-1
- d)246 g mol-1
Correct answer is option 'D'. Can you explain this answer?
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Given:
- A 5.25% solution of a substance is isotonic with a 21.5% solution of urea (NH2CONH2) by weight of solution in the same solvent at 300 K.
- The densities of both the solutions are assumed to be equal to 1.0 g cm-3.
To find: Molar mass of the substance
Solution:
Isotonic solutions are those that have the same osmotic pressure, which means that they have the same concentration of solute particles per unit volume. In this case, the 5.25% solution of the substance and the 21.5% solution of urea have the same osmotic pressure.
Step 1: Calculate the mole fraction of urea in the 21.5% solution.
The weight percentage of urea in the solution is 21.5%, which means that 100 g of the solution contains 21.5 g of urea and 78.5 g of the solvent. The number of moles of urea can be calculated as:
n(urea) = (21.5 g)/(60.06 g/mol) = 0.358 mol
The number of moles of solvent (water) can be calculated as:
n(solvent) = (78.5 g)/(18.015 g/mol) = 4.36 mol
The mole fraction of urea in the solution is:
X(urea) = n(urea)/(n(urea) + n(solvent)) = 0.076
Step 2: Calculate the mole fraction of the substance in the 5.25% solution.
Since the two solutions are isotonic, they have the same concentration of solute particles per unit volume. Therefore, the mole fraction of the substance in the 5.25% solution should be the same as the mole fraction of urea in the 21.5% solution. Thus,
X(substance) = X(urea) = 0.076
Step 3: Calculate the molecular weight of the substance.
The molecular weight of the substance can be calculated as:
M(substance) = (100 g)/(5.25 g/mol)/(6.02 x 10^23 molecules/mol) = 246 g/mol
Therefore, the correct answer is option D, 246 g/mol.
- A 5.25% solution of a substance is isotonic with a 21.5% solution of urea (NH2CONH2) by weight of solution in the same solvent at 300 K.
- The densities of both the solutions are assumed to be equal to 1.0 g cm-3.
To find: Molar mass of the substance
Solution:
Isotonic solutions are those that have the same osmotic pressure, which means that they have the same concentration of solute particles per unit volume. In this case, the 5.25% solution of the substance and the 21.5% solution of urea have the same osmotic pressure.
Step 1: Calculate the mole fraction of urea in the 21.5% solution.
The weight percentage of urea in the solution is 21.5%, which means that 100 g of the solution contains 21.5 g of urea and 78.5 g of the solvent. The number of moles of urea can be calculated as:
n(urea) = (21.5 g)/(60.06 g/mol) = 0.358 mol
The number of moles of solvent (water) can be calculated as:
n(solvent) = (78.5 g)/(18.015 g/mol) = 4.36 mol
The mole fraction of urea in the solution is:
X(urea) = n(urea)/(n(urea) + n(solvent)) = 0.076
Step 2: Calculate the mole fraction of the substance in the 5.25% solution.
Since the two solutions are isotonic, they have the same concentration of solute particles per unit volume. Therefore, the mole fraction of the substance in the 5.25% solution should be the same as the mole fraction of urea in the 21.5% solution. Thus,
X(substance) = X(urea) = 0.076
Step 3: Calculate the molecular weight of the substance.
The molecular weight of the substance can be calculated as:
M(substance) = (100 g)/(5.25 g/mol)/(6.02 x 10^23 molecules/mol) = 246 g/mol
Therefore, the correct answer is option D, 246 g/mol.
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5.25% solution of a substance is isotonic with a 21 .5% solution of urea (NH2CONH2 )by weight of solution in the same solvent at 300 K. If the densities of both the solutions are assumed to be equal to 1.0 g cm-3 then, molar mass of the substance will bea)90g mol-1b)115g mol-1c)105g mol-1d)246 g mol-1Correct answer is option 'D'. Can you explain this answer?
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5.25% solution of a substance is isotonic with a 21 .5% solution of urea (NH2CONH2 )by weight of solution in the same solvent at 300 K. If the densities of both the solutions are assumed to be equal to 1.0 g cm-3 then, molar mass of the substance will bea)90g mol-1b)115g mol-1c)105g mol-1d)246 g 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 5.25% solution of a substance is isotonic with a 21 .5% solution of urea (NH2CONH2 )by weight of solution in the same solvent at 300 K. If the densities of both the solutions are assumed to be equal to 1.0 g cm-3 then, molar mass of the substance will bea)90g mol-1b)115g mol-1c)105g mol-1d)246 g 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 5.25% solution of a substance is isotonic with a 21 .5% solution of urea (NH2CONH2 )by weight of solution in the same solvent at 300 K. If the densities of both the solutions are assumed to be equal to 1.0 g cm-3 then, molar mass of the substance will bea)90g mol-1b)115g mol-1c)105g mol-1d)246 g mol-1Correct answer is option 'D'. Can you explain this answer?.
5.25% solution of a substance is isotonic with a 21 .5% solution of urea (NH2CONH2 )by weight of solution in the same solvent at 300 K. If the densities of both the solutions are assumed to be equal to 1.0 g cm-3 then, molar mass of the substance will bea)90g mol-1b)115g mol-1c)105g mol-1d)246 g 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 5.25% solution of a substance is isotonic with a 21 .5% solution of urea (NH2CONH2 )by weight of solution in the same solvent at 300 K. If the densities of both the solutions are assumed to be equal to 1.0 g cm-3 then, molar mass of the substance will bea)90g mol-1b)115g mol-1c)105g mol-1d)246 g 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 5.25% solution of a substance is isotonic with a 21 .5% solution of urea (NH2CONH2 )by weight of solution in the same solvent at 300 K. If the densities of both the solutions are assumed to be equal to 1.0 g cm-3 then, molar mass of the substance will bea)90g mol-1b)115g mol-1c)105g mol-1d)246 g mol-1Correct answer is option 'D'. Can you explain this answer?.
Solutions for 5.25% solution of a substance is isotonic with a 21 .5% solution of urea (NH2CONH2 )by weight of solution in the same solvent at 300 K. If the densities of both the solutions are assumed to be equal to 1.0 g cm-3 then, molar mass of the substance will bea)90g mol-1b)115g mol-1c)105g mol-1d)246 g 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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