Answer to Question #156315 in General Chemistry for Bisal

Question

The first month of the winter in Bloomington, is a cold month, with an average temperature ranging between max. 5.3 °C and min. -4.1 °C. During this time there are a lot of snows and the roads are covered with ice. Which chemical compound(s) you use to lower the freezing point of ice for the streets of Bloomington and why? Suppose, you dissolve 87.0 g of rock salt in 500.0 g of water. What will be the freezing point of that solution? (Kf = 1.86 oC/molal for water)

Answer 1

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

ΔT = iK_fm

where

ΔT = Change in temperature in °C

i = van't Hoff factor

K_f = molal freezing point depression constant or cryoscopic constant = 1.86°C/molal for water

m = molality of the solute

number of moles of NaCl = $\dfrac{\textsf{mass}}{\textsf{molar mass}}$ = $\dfrac{87.0g}{58.4g/mol}$ = 1.5 moles

molality (m) of NaCl = $\dfrac{\textsf{moles of NaCl}}{\textsf{mass of water (kg)}}$ = $\dfrac{1.5}{500/1000}$ = $\dfrac{1.5}{2}$ = 0.75mol/kg

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, rock salt completely dissociates into the two ions, Na⁺ and Cl^-. Therefore, i = 2.

ΔT = iK_fm

∆T = 2 × 1.86 × 0.75 = 2.79°C.