Find the molarity of 15 solute of sulphuric acid
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heyy dear ur answer...
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
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 MMolarity of given solution is 1.7 M ……
hope this will help u..........@kundan
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
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 MMolarity of given solution is 1.7 M ……
hope this will help u..........@kundan
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