15% of Solution of H_{2}SO_{4} means 15 g of H_{2}SO_{4} are present in 100 g of the Solution i.e.

Mass of H_{2}SO_{4} dissolved = 15 g

Mass of the Solution = 100 g

Density of the Solution = 1.02 g/cm^{3} (given)

**Calculation of molality: **

Mass of Solution = 100 g ; Mass of H_{2}SO_{4} = 15g

Mass of water (solvent) = 100 – 15 = 85 g

Mol. mass of H_{2}SO_{4} = 98

∴ 15 g H_{2}SO_{4} = 15 / 98 = 0.153 moles

Thus 85 g of the solvent contain 0.153 moles 1000 g of the solvent contain

Hence the molality of H_{2}SO_{4} Solution = 1.8 m

**Calculation of molarity: **

15 g of H_{2}SO_{4} = 0.153 moles

Volume of Solution = mass/density = 100 / 1.02 = 98.04 mL

Thus 98.04 cm^{3 }of Solution contain H_{2}SO_{4} = 0.153 moles

1000 mL Solution contain

Hence molarity of the Solution is 1.56