Chemistry help!!

Question

NaCl is added to an ice cube tray with 0.25 kg of water and they don't freeze until they reach -4.5 degrees Celsius. What mass of NaCl was added to the ice cube tray?

Answer 1

∆T = imK

∆T = change in freezing point = 4.5 degrees

i = van't Hoff factor = 2 for NaCl

m = molality = moles NaCl/kg solvent

K = freezing point constant = 1.86

Solving for m we have m = ∆T/(i)(K) = 4.5/(2)(1.86) = 1.21 molal

1.21 m = 1.21 moles/kg x 0.25 kg = 0.303 moles NaCl

mass NaCl = 0.303 moles x 58.4 g/mole = 17.7 grams NaCl