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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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