Answer:
[tex]\boxed{\text{-3.6 $^{\circ}$C}}[/tex]
Explanation:
The formula for the freezing point depression ΔT_f by an electrolyte is
[tex]\Delta T_{f} = iK_{f}b[/tex]
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
[tex]n = \text{54.1 g} \times \dfrac{\text{1 mol}}{\text{56.11 g}} = \text{0.9642 mol}[/tex]
(b) Calculate the mass of water
You haven't given the density of the solution. It is about 1.04 g/mL
[tex]\text{Mass of solution} = \text{1000 mL} \times \dfrac{\text{1.04 g}}{\text{1 mL}} = \text{1040 g}[/tex]
Mass of water = 1040 g - 54.1 g = 986 g = 0.986 kg
(c) Calculate the molal concentration
[tex]b = \dfrac{\text{moles of solute}}{\text{kilograms of solvent}} = \dfrac{\text{0.9642 mol}}{\text{0.986 kg}} = \text{0.978 mol/kg}[/tex]
2. Calculate the freezing point
(a) Freezing point depression
[tex]\Delta T_{f} = iK_{f}b = 2 \times 1.86 \times 0.978 = 3.64 \,^{\circ}\text{C}[/tex]
(b) Freezing point
[tex]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}}$}[/tex]
