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Calculate the molality of 12.5% w/w sulphuric acid?
- a)2.85m
- b)3.15m
- c)1.45m
- d)2.50m
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Calculate the molality of 12.5% w/w sulphuric acid?a)2.85mb)3.15mc)1.4...
12.5% w/w means 12.5 g in 100 g of solution.
Weight of solvent = 100 g – 12.5 g= 87.5 g. Number of moles of sulphuric acid = 12.5/ 98 =0.127 mol
So molality = 0.127x1000/87.5 = 1.45 m
Most Upvoted Answer
Calculate the molality of 12.5% w/w sulphuric acid?a)2.85mb)3.15mc)1.4...
12.5%w/w means 12.5 gram of sulphuric acid is present in 100 gram of the solvent. now find molality by dividing total number of mole of sulphuric acid by Total weight of the solvent in kilogram
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Community Answer
Calculate the molality of 12.5% w/w sulphuric acid?a)2.85mb)3.15mc)1.4...
Calculation of Molality of Sulphuric Acid
To calculate the molality of a solution, we need to know the mass of the solute (sulphuric acid) and the mass of the solvent (water). The formula for molality is given by:
Molality (m) = moles of solute / mass of solvent (in kg)
Given Data:
- Percentage by weight of sulphuric acid = 12.5%
- Molar mass of sulphuric acid (H2SO4) = 98 g/mol
- Density of 12.5% w/w sulphuric acid = 1.10 g/mL
Calculation:
- Let's assume we have 100g of the solution.
- Mass of sulphuric acid in 100g of solution = 12.5g
- Number of moles of sulphuric acid = 12.5g / 98g/mol = 0.1276 mol
- Mass of water in 100g of solution = 100g - 12.5g = 87.5g
- Volume of the solution = 100g / 1.10 g/mL = 90.91 mL = 0.09091 L
- Mass of water = Volume x Density = 0.09091 L x 1000g/L = 90.91g = 0.09091 kg
Now, we can calculate the molality:
Molality (m) = 0.1276 mol / 0.09091 kg ≈ 1.45 m
Therefore, the molality of 12.5% w/w sulphuric acid is approximately 1.45m.
To calculate the molality of a solution, we need to know the mass of the solute (sulphuric acid) and the mass of the solvent (water). The formula for molality is given by:
Molality (m) = moles of solute / mass of solvent (in kg)
Given Data:
- Percentage by weight of sulphuric acid = 12.5%
- Molar mass of sulphuric acid (H2SO4) = 98 g/mol
- Density of 12.5% w/w sulphuric acid = 1.10 g/mL
Calculation:
- Let's assume we have 100g of the solution.
- Mass of sulphuric acid in 100g of solution = 12.5g
- Number of moles of sulphuric acid = 12.5g / 98g/mol = 0.1276 mol
- Mass of water in 100g of solution = 100g - 12.5g = 87.5g
- Volume of the solution = 100g / 1.10 g/mL = 90.91 mL = 0.09091 L
- Mass of water = Volume x Density = 0.09091 L x 1000g/L = 90.91g = 0.09091 kg
Now, we can calculate the molality:
Molality (m) = 0.1276 mol / 0.09091 kg ≈ 1.45 m
Therefore, the molality of 12.5% w/w sulphuric acid is approximately 1.45m.
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Calculate the molality of 12.5% w/w sulphuric acid?a)2.85mb)3.15mc)1.45md)2.50mCorrect answer is option 'C'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Calculate the molality of 12.5% w/w sulphuric acid?a)2.85mb)3.15mc)1.45md)2.50mCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Calculate the molality of 12.5% w/w sulphuric acid?a)2.85mb)3.15mc)1.45md)2.50mCorrect answer is option 'C'. Can you explain this answer?.
Calculate the molality of 12.5% w/w sulphuric acid?a)2.85mb)3.15mc)1.45md)2.50mCorrect answer is option 'C'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Calculate the molality of 12.5% w/w sulphuric acid?a)2.85mb)3.15mc)1.45md)2.50mCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Calculate the molality of 12.5% w/w sulphuric acid?a)2.85mb)3.15mc)1.45md)2.50mCorrect answer is option 'C'. Can you explain this answer?.
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