The density of 3 M solution of NaCl is 1.0 g mL⁻¹. Molality of the solution is ____________ × 10⁻² m. (Nearest integer).
Given: Molar mass of Na and Cl is 23 and 35.5 g mol⁻¹ respectively.

Solution

Chemistry ma’am · 4 steps

Take one litre of solution, subtract the salt's mass, and what is left is the water you divide by.

1. Pick one litre. Take 1 L = 1000 mL of solution. With density 1.0 g mL⁻¹, its total mass is 1000 g, and it holds 3 mol of NaCl.
2. Mass of solute. The molar mass of NaCl is 23 + 35.5 = 58.5 g mol⁻¹. So 3 mol weighs 3 × 58.5 = 175.5 g.
3. Mass of solvent. Molality needs only the solvent's mass. Water = 1000 - 175.5 = 824.5 g = 0.8245 kg.

   m = (3 mol)/(0.8245 kg) = 3.639 mol kg⁻¹

4. So. m = 363.9 × 10⁻² m, which rounds to 364.

The common slip: 300. Taking the whole 1000 g of solution as the solvent gives a molality of 3, that is 300, not 364. Subtract the 175.5 g of salt first.

The idea to keep. To go from molarity to molality, fix the volume at 1 L, use density for the total mass, and remove the solute's mass. Molality always divides by the solvent's mass alone.

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