# Find the molarity and molality of a 15% Solution of H2SO4 (density of H2SO4 = 1.020 g cm^−3) (Atomic mass: H = 1, O = 16, S = 32 amu).

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Find the molarity and molality of a 15% Solution of H2SO4 (density of H2SO4 = 1.020 g cm−3) (Atomic mass: H = 1, O = 16, S = 32 amu).

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15% of Solution of H2SO4 means 15 g of H2SO4 are present in 100 g of the Solution i.e.

Mass of H2SO4 dissolved = 15 g

Mass of the Solution = 100 g

Density of the Solution = 1.02 g/cm3 (given)

Calculation of molality:

Mass of Solution = 100 g ; Mass of H2SO4 = 15g

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

Mol. mass of H2SO4 = 98

∴ 15 g H2SO4 = 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 H2SO4 Solution = 1.8 m

Calculation of molarity:

15 g of H2SO4 = 0.153 moles

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

Thus 98.04 cm3 of Solution contain H2SO4 = 0.153 moles

1000 mL Solution contain Hence molarity of the Solution is 1.56