Molality (m) of 3 M aqueous solution of NaCl is : (Given : Density of solution =1.25 g mL⁻¹, Molar mass in g mol⁻¹: Na-23, Cl-35.5)

Solution

Chemistry ma’am · 4 steps

Take exactly 1 L of solution, then subtract the salt's mass to find how much water you really have.

1. Pick 1 L. Work with 1000 mL of solution. Its mass is 1000 × 1.25 = 1250 g, and it holds 3 mol of NaCl.
2. Mass of salt. Molar mass of NaCl is 23 + 35.5 = 58.5 g mol⁻¹, so the salt weighs 3 × 58.5 = 175.5 g.
3. Mass of water. Molality needs the mass of solvent only: 1250 - 175.5 = 1074.5 g = 1.0745 kg.

   m = (3 mol)/(1.0745 kg) = 2.79 mol kg⁻¹

4. So. The molality is 2.79 m, which is option (d).

The idea to keep. To turn molarity into molality, use density to get the solution's mass, then remove the solute's mass. Molality always divides by kg of solvent, never by kg of solution.

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