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damaskus [11]
3 years ago
15

The normal freezing point of a certain liquid Xis-7.30°C but when l02. g of iron(III) chloride (FeCl3) are dissolved in 650. g o

f Xthe solution freezes at -9.9°C instead. Use this information to calculate the molal freezing point depression constant Kf of X.
Chemistry
1 answer:
IRISSAK [1]3 years ago
7 0

Answer:

2.7 °C.kg/mol

Explanation:

Step 1: Calculate the freezing point depression (ΔT)

The normal freezing point of a certain liquid X is-7.30°C and the solution freezes at -9.9°C instead. The freezing point depression is:

ΔT = -7.30 °C - (-9.9 °C) = 2.6 °C

Step 2: Calculate the molality of the solution (b)

We will use the following expression.

b = mass of solute / molar mass of solute × kilograms of solvent

b = 102. g / (162.2 g/mol) × 0.650 kg = 0.967 mol/kg

Step 3: Calculate the molal freezing point depression constant Kf of X

Freezing point depression is a colligative property. It can be calculated using the following expression.

ΔT = Kf × b

Kf = ΔT / b

Kf = 2.6 °C / (0.967 mol/kg) = 2.7 °C.kg/mol

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Considering the ICE table for the equilibrium as:

\begin{matrix} & 2SO_3_{(g)} & \rightleftharpoons & 2SO_2_{(g)} & + & O_2_{(g)}\\At\ time, t = 0 & 0.5067 &&0&&0 \\ At\ time, t=t_{eq} & -2x &&2x&&x \\----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:- &0.5067-2x&&2x&&x\end{matrix}

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The expression for the equilibrium constant is:

K_c=\frac {[SO_2]^2[O_2]}{[SO_3]^2}  

K_c=\frac{(0.3334)^2\times 0.0867}{(0.3334)^2}

K_c=0.0867

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