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On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g residue is formed. Thus, molality of solution is A. 2000 molal B. 1.825 molal C. 1.000 molal D. 1.200 molal?
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On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g resid...
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On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g resid...
Given data:
- Volume of urea solution = 1 L
- Density of urea solution = 1.2 g/mL
- Mass of residue formed = 120 g
Calculating the mass of the solution:
Mass = Density × Volume
Mass of solution = 1.2 g/mL × 1000 mL = 1200 g
Calculating the mass of urea:
Mass of urea = Mass of solution - Mass of residue
Mass of urea = 1200 g - 120 g = 1080 g
Calculating the number of moles of urea:
Molar mass of urea (CH4N2O) = 12.01 g/mol + 4(1.01 g/mol) + 2(14.01 g/mol) + 16.00 g/mol = 60.06 g/mol
Number of moles of urea = Mass of urea / Molar mass of urea
Number of moles of urea = 1080 g / 60.06 g/mol = 17.97 mol
Calculating the molality of the solution:
Molality (m) = Number of moles of solute / Mass of solvent in kg
Since the solvent is water, its mass is equal to its volume.
Mass of solvent = Volume of urea solution × Density of urea solution
Mass of solvent = 1 L × 1.2 g/mL = 1200 g = 1.2 kg
Molality (m) = 17.97 mol / 1.2 kg = 14.975 mol/kg ≈ 1.825 molal
Answer:
The molality of the urea solution is approximately 1.825 molal. Therefore, the correct option is B. 1.825 molal.
- Volume of urea solution = 1 L
- Density of urea solution = 1.2 g/mL
- Mass of residue formed = 120 g
Calculating the mass of the solution:
Mass = Density × Volume
Mass of solution = 1.2 g/mL × 1000 mL = 1200 g
Calculating the mass of urea:
Mass of urea = Mass of solution - Mass of residue
Mass of urea = 1200 g - 120 g = 1080 g
Calculating the number of moles of urea:
Molar mass of urea (CH4N2O) = 12.01 g/mol + 4(1.01 g/mol) + 2(14.01 g/mol) + 16.00 g/mol = 60.06 g/mol
Number of moles of urea = Mass of urea / Molar mass of urea
Number of moles of urea = 1080 g / 60.06 g/mol = 17.97 mol
Calculating the molality of the solution:
Molality (m) = Number of moles of solute / Mass of solvent in kg
Since the solvent is water, its mass is equal to its volume.
Mass of solvent = Volume of urea solution × Density of urea solution
Mass of solvent = 1 L × 1.2 g/mL = 1200 g = 1.2 kg
Molality (m) = 17.97 mol / 1.2 kg = 14.975 mol/kg ≈ 1.825 molal
Answer:
The molality of the urea solution is approximately 1.825 molal. Therefore, the correct option is B. 1.825 molal.
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On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g residue is formed. Thus, molality of solution is A. 2000 molal B. 1.825 molal C. 1.000 molal D. 1.200 molal?
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On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g residue is formed. Thus, molality of solution is A. 2000 molal B. 1.825 molal C. 1.000 molal D. 1.200 molal? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g residue is formed. Thus, molality of solution is A. 2000 molal B. 1.825 molal C. 1.000 molal D. 1.200 molal? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g residue is formed. Thus, molality of solution is A. 2000 molal B. 1.825 molal C. 1.000 molal D. 1.200 molal?.
On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g residue is formed. Thus, molality of solution is A. 2000 molal B. 1.825 molal C. 1.000 molal D. 1.200 molal? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g residue is formed. Thus, molality of solution is A. 2000 molal B. 1.825 molal C. 1.000 molal D. 1.200 molal? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g residue is formed. Thus, molality of solution is A. 2000 molal B. 1.825 molal C. 1.000 molal D. 1.200 molal?.
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