Find the molarity and molality of 15 solution of h2so4
Answers
Explanation:
No. of moles of H₂SO₄ = 15 g / 98 g/mol = 0.153 mol
Weight of solvent = 100g - 15g = 85g = 85 × 10^-3 kg
Molality = Moles of solute / Weight of Solvent in kg
= 0.153 / (85 × 10^-3)
= 1.8 m
Molality of given solution is 1.8 m ……[1]
Volume of solution = Mass of solution / Density
= 100 g / 1.1 g/cm^3
= 90.90 cm^3
= 0.090 Litres
Molarity = Moles of solute / Volume of solution in litres
= 0.153 / 0.090
≈ 1.7 M
Molarity of given solution is 1.7 M ……[2]
15% H₂SO₄ solution means 15 g of H₂SO₄ is present in 100 g of Water.
No. of moles of H₂SO₄ = 15 g / 98 g/mol = 0.153 mol
Weight of solvent = 100g - 15g = 85g = 85 × 10^-3 kg
Molality = Moles of solute / Weight of Solvent in kg
= 0.153 / (85 × 10^-3)
= 1.8 m
Molality of given solution is 1.8 m ……[1]
Volume of solution = Mass of solution / Density
= 100 g / 1.1 g/cm^3
= 90.90 cm^3
= 0.090 Litres
Molarity = Moles of solute / Volume of solution in litres
= 0.153 / 0.090
≈ 1.7 M
Molarity of given solution is 1.7 M ……[2]