Solutions Question 6
Question: Calculate the molality of 1 litre solution of 93% $ H_2SO_4 $ (weight/volume). The density of the solution is 1.84 g /ml [UPSEAT 2000]
Options:
A) 10.43
B) 20.36
C) 12.05
D) 14.05
Show Answer
Answer:
Correct Answer: A
Solution:
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First, let’s understand what we’re given:
- 1 litre solution of 93% H₂SO₄ (w/v)
- Density of solution = 1.84 g/ml
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Calculate the mass of the solution: Mass of solution = Density × Volume = 1.84 g/ml × 1000 ml = 1840 g
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Calculate the mass of H₂SO₄ in the solution: 93% w/v means 93 g of H₂SO₄ in 100 ml of solution In 1000 ml (1 L), mass of H₂SO₄ = 93 × 10 = 930 g
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Calculate the mass of water: Mass of water = Mass of solution - Mass of H₂SO₄ = 1840 g - 930 g = 910 g
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Convert mass of water to kg: 910 g = 0.910 kg
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Calculate the number of moles of H₂SO₄: Molar mass of H₂SO₄ = 98.08 g/mol Moles of H₂SO₄ = 930 g ÷ 98.08 g/mol = 9.482 mol
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Calculate molality: Molality = moles of solute / kg of solvent = 9.482 mol / 0.910 kg = 10.42 mol/kg
BETA
