15% H₂SO₄ Solution | Molarity and Molality Trick | JEE Main PYQ

 

❓ Question

A 15% (w/w) H₂SO₄ solution has a density of 1.020 g cm⁻³.

Find:

1. Molarity (M)
2. Molality (m)

(Given: Molar mass of H₂SO₄ = 98 g mol⁻¹)

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✍️ Solution

Assume 100 g of solution.

→ Mass of H₂SO₄ (solute) = 15 g

→ Mass of water (solvent) = 100 − 15 = 85 g = 0.085 kg

Given,

$$\text{Density} = 1.020\ \text{g cm}^{-3}$$

Volume of solution,

$$V=\frac{\text{Mass}}{\text{Density}} =\frac{100}{1.020} =98.04\ \text{mL} =0.09804\ \text{L}$$

Number of moles of H₂SO₄,

$$n=\frac{15}{98}=0.1531\ \text{mol}$$

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1. Molarity

$$M=\frac{\text{Moles of solute}}{\text{Volume of solution (L)}}$$
$$=\frac{0.1531}{0.09804} \approx1.56\ \text{M}$$

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2. Molality

$$m=\frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}}$$
$$=\frac{0.1531}{0.085} \approx1.80\ \text{m}$$

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✅ Final Answer

→ Molarity = $\boxed{1.56\ \text{M}}$

→ Molality = $\boxed{1.80\ \text{m}}$