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?
Answers
Answered by
15
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
Similar questions