JEE Challenger
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ChemMedAdv 2019 • Paper 1

Freezing Point Depression of Benzene Solution

Upon dissolving 0.5 g0.5\text{ g} of a non-volatile non-ionic solute into 39 g39\text{ g} of benzene, the vapor pressure drops from 650 mm Hg650\text{ mm Hg} to 640 mm Hg640\text{ mm Hg}. Calculate the depression in freezing point of benzene (in K) caused by adding this solute. (Data provided: The molar mass of benzene is 78 g mol−178\text{ g mol}^{-1} and its molal freezing point depression constant is 5.12 K kg mol−15.12\text{ K kg mol}^{-1})

Official Numerical Answer1.02 to 1.03

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Step-by-Step Solution

To calculate the depression in freezing point (ΔTf\Delta T_f) of benzene, we first determine the number of moles of the solute using Raoult's law for non-volatile solutes.

Step 1: Application of Raoult's Law

According to Raoult's law for relative lowering of vapor pressure: P∘−PP∘=X2=n2n1+n2\frac{P^\circ - P}{P^\circ} = X_2 = \frac{n_2}{n_1 + n_2}

where:

  • P∘=650 mm HgP^\circ = 650\text{ mm Hg} (Vapor pressure of pure benzene)
  • P=640 mm HgP = 640\text{ mm Hg} (Vapor pressure of the solution)
  • n1n_1 is the number of moles of benzene (solvent)
  • n2n_2 is the number of moles of the solute

Rearranging the equation yields: P∘−PP=n2n1\frac{P^\circ - P}{P} = \frac{n_2}{n_1}

Step 2: Calculate the Moles of Benzene (n1n_1) and Solute (n2n_2)

Given:

  • Mass of benzene, w1=39 gw_1 = 39\text{ g}
  • Molar mass of benzene, M1=78 g mol−1M_1 = 78\text{ g mol}^{-1}

n1=w1M1=39 g78 g mol−1=0.5 moln_1 = \frac{w_1}{M_1} = \frac{39\text{ g}}{78\text{ g mol}^{-1}} = 0.5\text{ mol}

Now, substitute n1n_1 into the modified Raoult's law equation: 650−640640=n20.5\frac{650 - 640}{640} = \frac{n_2}{0.5} 10640=n20.5\frac{10}{640} = \frac{n_2}{0.5} n2=10×0.5640=5640=1128 moln_2 = \frac{10 \times 0.5}{640} = \frac{5}{640} = \frac{1}{128}\text{ mol}

Step 3: Calculate the Molality (mm) of the Solution

Molality is defined as the moles of solute per kilogram of solvent: m=n2Mass of solvent in kg=1128391000=1000128×39 mol kg−1m = \frac{n_2}{\text{Mass of solvent in kg}} = \frac{\frac{1}{128}}{\frac{39}{1000}} = \frac{1000}{128 \times 39}\text{ mol kg}^{-1}

Step 4: Calculate the Depression in Freezing Point (ΔTf\Delta T_f)

The formula for depression in freezing point is: ΔTf=Kf×m\Delta T_f = K_f \times m

Given Kf=5.12 K kg mol−1K_f = 5.12\text{ K kg mol}^{-1}: ΔTf=5.12×1000128×39\Delta T_f = 5.12 \times \frac{1000}{128 \times 39}

Since 5.12128=0.04\frac{5.12}{128} = 0.04: ΔTf=0.04×100039=4039 K≈1.0256 K\Delta T_f = 0.04 \times \frac{1000}{39} = \frac{40}{39}\text{ K} \approx 1.0256\text{ K}

Rounding off to two decimal places, we get: ΔTf≈1.03 K(or 1.02 K)\Delta T_f \approx 1.03\text{ K} \quad (\text{or } 1.02\text{ K})

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