Molality of 0.8 M H₂SO₄ solution (density 1.06 g cm⁻³) is ________ ×10⁻³ m.

Solution

Chemistry ma’am · 4 steps

Take 1 L of solution, find its mass from density, then remove the acid's mass to get the solvent.

1. Pick 1 L. Work with 1 L = 1000 cm³ of solution. At 1.06 g cm⁻³, its mass is 1000 × 1.06 = 1060 g.
2. Mass of acid. 0.8 M means 0.8 mol of H₂SO₄ in this litre. With molar mass 98 g mol⁻¹, that is 0.8 × 98 = 78.4 g.
3. Mass of water. Molality needs the mass of solvent, not solution. Water = 1060 - 78.4 = 981.6 g = 0.9816 kg.

   m = (0.8 mol)/(0.9816 kg) = 0.815 mol kg⁻¹

4. So. 0.815 m = 815 × 10⁻³ m, so the answer is 815.

The common slip: 755. Dividing by the whole solution's mass, 1.06 kg, gives 0.8/1.06 = 0.755 m, that is 755, not 815. The acid's 78.4 g must come out first.

The idea to keep. To go from molarity to molality, fix 1 L of solution, get its mass from density, and subtract the solute's mass. Molality always divides by kilograms of solvent only.

Answer from the NTA final answer key. Solution by PrepWiser.