Determine the mass of NaCl necessary to lower the freezing point of 300 mL of distilled water to -10C?

Freezing point depression, ΔT = 10°C
For water, the cryoscopic constant, K = 1.853 kg°C/mol
Molality of ions, m = ? mol/kg

ΔT = K m
10°C = (1.853 kg°C/mol) × m
m = 10/1.853 mol/kg

1 mole of NaCl gives 2 moles of ions (1 mole of Na⁺ ions and 1 mole of Cl⁻ ions).
Molality of NaCl = (10/1.853 mol/kg) × (1/2) = 5/1.853 mol/kg

Mass of water = Density × Volume = (1.00 g/mL) × (300 mL) = 300 g = 0.3 kg
No. of moles of NaCl = (5/1.853 mol/kg) × (0.3 kg) = 1.5/1.853 mol

Molar mass of NaCl = (23.0 + 35.5) g/mol = 58.5 g/mol
Mass of NaCl = (1.5/1.853 mol) × (58.5 g/mol) = 47.4 g (3 sig. fig.)