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Which of the given aqueous solution has maximum boiling point?
(Kb of H20 = 0 .5 2 ° mol-1 kg), Tb (H20 ) = 373 K)
- a)18% glucose solution by weight of solution
- b)34.2% sucrose solution by weight of solution
- c)6% urea solution by weight of solution
- d)All have equal boiling points
Correct answer is option 'B'. Can you explain this answer?
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Which of the given aqueous solution has maximum boiling point?(Kb of H...
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Which of the given aqueous solution has maximum boiling point?(Kb of H...
Explanation:
The boiling point of a solution depends on the concentration of solute particles in the solution. The greater the concentration of solute particles, the higher the boiling point of the solution. The boiling point elevation is given by the equation ΔTb = Kb × m, where ΔTb is the boiling point elevation, Kb is the molal boiling point constant, and m is the molality of the solute.
Calculation:
To determine which aqueous solution has the maximum boiling point, we need to calculate the molality of the solute in each solution.
a) 18% glucose solution by weight of solution:
This means that 18 g of glucose is present in 100 g of solution.
Molar mass of glucose (C6H12O6) = 180 g/mol
Number of moles of glucose = 18 g / 180 g/mol = 0.1 mol
Mass of water = 100 g - 18 g = 82 g
Molar mass of water (H2O) = 18 g/mol
Number of moles of water = 82 g / 18 g/mol = 4.56 mol
Molality of glucose = 0.1 mol / 4.56 kg = 0.022 mol/kg
b) 34.2% sucrose solution by weight of solution:
This means that 34.2 g of sucrose is present in 100 g of solution.
Molar mass of sucrose (C12H22O11) = 342 g/mol
Number of moles of sucrose = 34.2 g / 342 g/mol = 0.1 mol
Mass of water = 100 g - 34.2 g = 65.8 g
Molar mass of water (H2O) = 18 g/mol
Number of moles of water = 65.8 g / 18 g/mol = 3.66 mol
Molality of sucrose = 0.1 mol / 3.66 kg = 0.027 mol/kg
c) 6% urea solution by weight of solution:
This means that 6 g of urea is present in 100 g of solution.
Molar mass of urea (CO(NH2)2) = 60 g/mol
Number of moles of urea = 6 g / 60 g/mol = 0.1 mol
Mass of water = 100 g - 6 g = 94 g
Molar mass of water (H2O) = 18 g/mol
Number of moles of water = 94 g / 18 g/mol = 5.22 mol
Molality of urea = 0.1 mol / 5.22 kg = 0.019 mol/kg
Comparison:
Comparing the molalities of the solutes in each solution:
- Glucose solution: 0.022 mol/kg
- Sucrose solution: 0.027 mol/kg
- Urea solution: 0.019 mol/kg
Conclusion:
The sucrose solution (34.2% sucrose solution by weight of solution) has the highest molality of solute particles and therefore the highest boiling point. Therefore, option B is the correct answer.
The boiling point of a solution depends on the concentration of solute particles in the solution. The greater the concentration of solute particles, the higher the boiling point of the solution. The boiling point elevation is given by the equation ΔTb = Kb × m, where ΔTb is the boiling point elevation, Kb is the molal boiling point constant, and m is the molality of the solute.
Calculation:
To determine which aqueous solution has the maximum boiling point, we need to calculate the molality of the solute in each solution.
a) 18% glucose solution by weight of solution:
This means that 18 g of glucose is present in 100 g of solution.
Molar mass of glucose (C6H12O6) = 180 g/mol
Number of moles of glucose = 18 g / 180 g/mol = 0.1 mol
Mass of water = 100 g - 18 g = 82 g
Molar mass of water (H2O) = 18 g/mol
Number of moles of water = 82 g / 18 g/mol = 4.56 mol
Molality of glucose = 0.1 mol / 4.56 kg = 0.022 mol/kg
b) 34.2% sucrose solution by weight of solution:
This means that 34.2 g of sucrose is present in 100 g of solution.
Molar mass of sucrose (C12H22O11) = 342 g/mol
Number of moles of sucrose = 34.2 g / 342 g/mol = 0.1 mol
Mass of water = 100 g - 34.2 g = 65.8 g
Molar mass of water (H2O) = 18 g/mol
Number of moles of water = 65.8 g / 18 g/mol = 3.66 mol
Molality of sucrose = 0.1 mol / 3.66 kg = 0.027 mol/kg
c) 6% urea solution by weight of solution:
This means that 6 g of urea is present in 100 g of solution.
Molar mass of urea (CO(NH2)2) = 60 g/mol
Number of moles of urea = 6 g / 60 g/mol = 0.1 mol
Mass of water = 100 g - 6 g = 94 g
Molar mass of water (H2O) = 18 g/mol
Number of moles of water = 94 g / 18 g/mol = 5.22 mol
Molality of urea = 0.1 mol / 5.22 kg = 0.019 mol/kg
Comparison:
Comparing the molalities of the solutes in each solution:
- Glucose solution: 0.022 mol/kg
- Sucrose solution: 0.027 mol/kg
- Urea solution: 0.019 mol/kg
Conclusion:
The sucrose solution (34.2% sucrose solution by weight of solution) has the highest molality of solute particles and therefore the highest boiling point. Therefore, option B is the correct answer.
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Which of the given aqueous solution has maximum boiling point?(Kb of H20 = 0 .5 2 ° mol-1 kg), Tb (H20 ) = 373 K)a)18% glucose solution by weight of solutionb)34.2% sucrose solution by weight of solutionc)6% urea solution by weight of solutiond)All have equal boiling pointsCorrect answer is option 'B'. Can you explain this answer?
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Solutions for Which of the given aqueous solution has maximum boiling point?(Kb of H20 = 0 .5 2 ° mol-1 kg), Tb (H20 ) = 373 K)a)18% glucose solution by weight of solutionb)34.2% sucrose solution by weight of solutionc)6% urea solution by weight of solutiond)All have equal boiling pointsCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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