Question

The molality of 15% (%w/vol) solution of H2SO4 of density 1.1g/cm^3 is approximately

Answer 1

Answer:1.61

Explanation:

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