How much amount of NaCl should be added to 600 g of water (ρ = 1.00 g/mL) to decrease the freezing point of water to – 0.2 °C ? ______. (Round off to the nearest integer)
(The freezing point depression constant for water = 2K kg mol –1 )

Solution

Chemistry ma’am · 4 steps

NaCl splits into two ions, so put i = 2 in Δ T_f = i K_f m before you solve for the molality.

1. Set up. Pure water freezes at 0 °C, so Δ T_f = 0.2 K. NaCl → Na⁺ + Cl⁻ gives two particles per formula unit, so i = 2.

   Δ T_f = i K_f m ⇒ m = 0.2/(2 × 2) = 0.05 mol kg⁻¹

2. Moles of NaCl. Molality uses kilograms of solvent, and 600 g of water is 0.6 kg. The density is not needed because the mass is already given. Moles = 0.05 × 0.6 = 0.03 mol.
3. Convert to mass. The molar mass of NaCl is 23 + 35.5 = 58.5 g mol⁻¹, so the mass is 0.03 × 58.5 = 1.755 g.
4. So. About 1.76 g of NaCl is needed, which rounds to 2.

The common slip: 4. Forgetting the van 't Hoff factor and taking i = 1 doubles the molality to 0.1 mol kg⁻¹ and gives 3.51 g, which rounds to 4.

The idea to keep. Colligative properties count particles, not formula units. For an electrolyte, find i first, then get the molality, and convert moles to grams at the end.

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