Question

An aqueous solution containing 58.8% w/v of

H2SO4 has density 1.388 g/mL. The molality
with respect to H2SO4 of the solution is

Answer 1

amount of solute is 58.8% ( w/v)

∴ 58.8g of solute (H₂SO₄) is present in 100 mL of solution.

But density of solution is 1.388 g/ml

Hence, mass of solution = volume of solution × density of solution

= 100mL × 1.388 g/mL

= 138.8g

∴ mass of solvent = mass of solution - mass of solute

= 138.8g - 58.8g = 80g

Now, molality = mole of solute × 1000/mass of solvent in g

= {weight of solute} × 1000/molecular mass of solute × mass of solvent

= 58.8 × 1000/98 × 80 [ ∵ molecular mass of H₂SO₄ = 98 g/mol]

= 7.5m

Hence, molality = 7.5m