Question

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).  

Answer 1

Mass of H₂SO₄ = 15g

Mass of solution = 100 g

d = 1.020 g/cm³

Molar mass of H₂SO₄ = (1 x 2) + (32 x 1) + (16 x 4) = 98 g/mol

No. of moles of H₂SO₄ = mass / molar mass

\(= \frac{15 g}{98 g/mol}\)

= 0.153 moles

Molarity

Volume of solution = mass / density

\(= \frac{100}{1.02}\)

= 98.04 ml

= 98.04 cm³

No. of moles of H₂SO₄ in 1000 ml = \(\frac{0.153}{98.04} \times 1000\)

= 1.56 moles

= 1.56 m

Molarity 

Mass of solvent = 100g - 15g = 85g

No. of moles of solvent in 1000g = \(\frac{0.153}{85} \times 1000\)

= 1.8 moles

= 1.8 m

Answer 2

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