What is the new freezing point

Question

What is the new freezing point if you put in 100.0 g of sodium chloride into 750. mL of water (density 1.00 g/mL)(

Answer 1

For this colligative property problem, you use the formula ∆T = imK

∆T = change in temperature = ?

i = van't Hoff factor = 2 for NaCl

m = molality = moles solute/kg solvent = 100.0g/58.4 g/mole/0.750 kg = 2.283 m

K = freezing point constant for water = -1.86 ºC/m

∆T = (2)(2.283)(1.86) = 8.49

New freezing point of solution = -8.49 ºC