The freezing point (in °C) of a solution containing 0.1 g of K3[Fe(CN)6] (Mol. Wt. 329) in 100 g of water (Kf = 1.86 K kg mol

Engineering

Chemistry

vant Hoff Factor

Colligative properties and Measurement

Concentration Terms

Question

The freezing point (in °C) of a solution containing 0.1 g of K₃[Fe(CN)₆] (Mol. Wt. 329) in 100 g of water (K_f = 1.86 K kg mol^–1) is

–5.7 × 10^–2

–1.2 × 10^–2

–2.3 × 10^–2

–5.7 × 10^–3

$∆$ T = k_f × m × i × 1000

$= 1.86 \times \frac{0.1}{329 \times 100} \times 4 \times 1000$

= 2.26 × 10^–2 $\approx$ 2.3 × 10^–2

This question involves calculating the freezing point depression of a solution using colligative properties. The key concept is that adding a solute lowers the freezing point of a solvent, and the amount of depression depends on the molality of the solution and the van't Hoff factor (i), which accounts for the dissociation of the solute.

Step 1: Calculate the molality (m) of the solution.

Molality (m) is defined as the number of moles of solute per kilogram of solvent.

Given:

Mass of solute, K₃[Fe(CN)₆] = 0.1 g
Molar mass of solute = 329 g mol^-1
Mass of solvent (water) = 100 g = 0.1 kg

First, find the number of moles of solute:

n = \frac{mass of solute}{molar mass} = \frac{0.1}{329} = 3.04 \times 10^{- 4} mol

Now, calculate molality:

m = \frac{n}{mass of solvent in kg} = \frac{3.04 \times 10^{- 4}}{0.1} = 3.04 \times 10^{- 3} mol kg^{- 1}

Step 2: Determine the van't Hoff factor (i).

K₃[Fe(CN)₆] is an ionic compound that dissociates in water:

K_{3} [Fe {(CN)}_{6}] \to 3 K^{+} + {[Fe {(CN)}_{6}]}^{3 -}

This produces 4 ions in total. Therefore, the van't Hoff factor i = 4.

Step 3: Apply the freezing point depression formula.

The formula for the depression in freezing point (ΔT_f) is:

Δ T_{f} = i \times K_{f} \times m

Given K_f = 1.86 K kg mol^-1, i = 4, m = 3.04 × 10^-3 mol kg^-1.

Δ T_{f} = 4 \times 1.86 \times 3.04 \times 10^{- 3}

First, calculate 4 × 1.86 = 7.44

Then, 7.44 × 3.04 × 10^-3 = 22.6176 × 10^-3 = 0.0226176 K

This is approximately 2.26 × 10^-2 K.

Since freezing point depression is a decrease, ΔT_f = -2.26 × 10^-2 °C (The magnitude in K and °C is the same).

Step 4: Compare with options.

The calculated value is -2.26 × 10^-2 °C, which is very close to the option -2.3 × 10^-2 °C (the slight difference is due to rounding during calculation).

Final Answer: The freezing point of the solution is -2.3 × 10^-2 °C.

Related Concepts and Formulae

Colligative Properties: These are properties of solutions that depend on the number of solute particles in a given amount of solvent, not on the nature of the solute particles. Freezing point depression is one such property.

Freezing Point Depression: The decrease in the freezing point of a solvent upon the addition of a non-volatile solute. The formula is:

Δ T_{f} = i \times K_{f} \times m

Where:

ΔT_f = Freezing point depression
i = van't Hoff factor (number of ions per formula unit for electrolytes)
K_f = Cryoscopic constant (molal freezing point depression constant) of the solvent
m = Molality of the solution (mol of solute per kg of solvent)

van't Hoff Factor (i): This factor indicates the number of particles a solute dissociates into in solution. For non-electrolytes, i = 1. For electrolytes, i equals the total number of ions produced per formula unit.

Molality (m): A concentration unit defined as:

m = \frac{moles of solute}{mass of solvent (in kg)}

Detailed Solution

Related Concepts and Formulae

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