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InterviewSolution
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The molality of 15% (wt/vol.) solution ofH2SO4 of density 1.1 g/cm3 isapproximately- |
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Answer» Amount of solute is 15% ( w/v) ∴ 15g of solute (H₂SO₄) is present in 100 mL of solution But density of solution is 1.1 g/cm³ Hence, mass of solution = volume of solution × density of solution= 100mL × 1.1 g/mL [ ∵ 1cm³ = 1 mL ] = 110g ∴ mass of solvent = mass of solution - mass of solute = 110g - 15g = 95g Now, molality = mole of solute × 1000/mass of solvent in g = {weight of solute} × 1000/molecular mass of solute × mass of solvent = 15 × 1000/98 × 95 [ ∵ molecular mass of H₂SO₄ = 98 g/mol= 1.61 Hence, molality = 1.61 |
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