Question

1. salt for de-icing roads

road crews spread rock salt on icy pavement, creating a solution equivalent to dissolving 50.0 g nacl in 2.00 kg of water on the road surface.

• i=2
• kf=1.86

what is the freezing point of the salty water? (pure water freezes at 0.0 °c.)

Explanation:

Step1: Calculate moles of NaCl

Molar mass of NaCl = 58.44 g/mol. Moles = $\frac{50.0\ \text{g}}{58.44\ \text{g/mol}} \approx 0.8556\ \text{mol}$

Step2: Find molality of solution

Molality ($m$) = $\frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{0.8556\ \text{mol}}{2.00\ \text{kg}} = 0.4278\ \text{mol/kg}$

Step3: Compute freezing point depression

$\Delta T_f = i \times K_f \times m = 2 \times 1.86\ ^\circ\text{C·kg/mol} \times 0.4278\ \text{mol/kg} \approx 1.59\ ^\circ\text{C}$

Step4: Determine freezing point of solution

Freezing point = $0.0\ ^\circ\text{C} - 1.59\ ^\circ\text{C} = -1.59\ ^\circ\text{C}$

Answer:

$-1.6\ ^\circ\text{C}$ (rounded to two significant figures)