Question

the molarity of 15% (wt/vol) solution of H2SO4 of density 1.1 g/cm3 is approximately​

Answer 1

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