...(ii) As molality =3.3mol∕kg that means 3.3 moles of KCl present in 1kg of solvent. ∴ 1000g of solvent has 3.3×74.5g of KCl=245.8g ∴ Weight of solution = weight of solute + Weight of solvent =245.85g+1000g=1245.85g From Eqs. (ii) Volume of solution =
Massofsolution
1.20g∕mL
=
1245.85g
1.20g∕mL
=1038.2mL From Eq. (i) Molarity =
245.85g
74.5g
×
1
1038.2mL
×1000=3.17mol∕L ∴ Molarity of solution in mol∕Lis3.17 Answer = 3