Question

What is the freezing point of a water solution with the total volume of 1.00L, that is made by dissolving 54.1 grams of potassium hydroxide, KOH in water? Assume the KOH dissociates completely into ions when it dissolves and Kf for water is 1.86 Celsius/M

Freezing point =

Answer 1

Answer:

$\boxed{\text{-3.6 ^{\circ} C}}$  

Explanation:

The formula for the freezing point depression ΔT_f by an electrolyte is

$\Delta T_{f} = iK_{f}b$

1. Calculate i

KOH(aq) ⟶ K⁺(aq) + OH⁻(aq)

1 mol KOH gives 2 mol ions.

i = 2

2. Calculate the molal concentration of KOH

(a) Moles of KOH

$n = \text{54.1 g} \times \dfrac{\text{1 mol}}{\text{56.11 g}} = \text{0.9642 mol}$

(b) Calculate the mass of water

You haven't given the density of the solution.  It is about 1.04 g/mL

$\text{Mass of solution} = \text{1000 mL} \times \dfrac{\text{1.04 g}}{\text{1 mL}} = \text{1040 g}$

Mass of water = 1040 g - 54.1 g = 986 g = 0.986 kg

(c) Calculate the molal concentration

$b = \dfrac{\text{moles of solute}}{\text{kilograms of solvent}} = \dfrac{\text{0.9642 mol}}{\text{0.986 kg}} = \text{0.978 mol/kg}$

2. Calculate the freezing point

(a) Freezing point depression

$\Delta T_{f} = iK_{f}b = 2 \times 1.86 \times 0.978 = 3.64 \,^{\circ}\text{C}$

(b) Freezing point

$T_{f} = T_{f}^{^\circ} - \Delta T_{f} = 0.00 - 3.64 = \textbf{-3.6 ^{\circ} C}\\\\\text{The freezing point of the solution is \large \boxed{\textbf{-3.6 ^{\circ} C}} }$