**Problem:**

31.65 g of sodium chloride is added to 220.0 mL of water at 34 °C. How will this affect the freezing point of the water?

Assume the sodium chloride completely dissociates in the water.

Given: density of water at 35 °C = 0.994 g/mL

K

_{f}water = 1.86 °C kg/mol

**Solution:**

To find the temperature change elevation of a solvent by a solute, use the equation:

ΔT = iK

_{f}m

where

ΔT = Change in temperature in °C

i = van 't Hoff factor

K

_{f}= molal freezing point depression constant or cryoscopic constant in °C kg/mol

m = molality of the solute in mol solute/kg solvent.

**Step 1**Calculate the molality of the NaCl

molality (m) of NaCl = moles of NaCl/kg water

From the periodic table

atomic mass Na = 22.99

atomic mass Cl = 35.45

moles of NaCl = 31.65 g x 1 mol/(22.99 + 35.45)

moles of NaCl = 31.65 g x 1 mol/58.44 g

moles of NaCl = 0.542 mol

kg water = density x volume

kg water = 0.994 g/mL x 220 mL x 1 kg/1000 g

kg water = 0.219 kg

m

_{NaCl}= moles of NaCl/kg water

m

_{NaCl}= 0.542 mol/0.219 kg

m

_{NaCl}= 2.477 mol/kg

**Step 2**Determine the van 't Hoff factor

The van 't Hoff factor, i, is a constant associated with the amount of dissociation of the solute in the solvent. For substances which do not dissociate in water, such as sugar, i = 1. For solutes that completely dissociate into two ions, i = 2. For this example NaCl completely dissociates into the two ions, Na

^{+}and Cl

^{-}. Therefore, i = 2 for this example.

**Step 3**Find ΔT

ΔT = iK

_{f}m

ΔT = 2 x 1.86 °C kg/mol x 2.477 mol/kg

ΔT = 9.21 °C

**Answer:**

Adding 31.65 g of NaCl to 220.0 mL of water will lower the freezing point by 9.21 °C.