JEE Challenger
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ChemMedAdv 2017 • Paper 2

Vapour Pressure versus Temperature Plot for Ethanol Solution Freezing Point

Pure water freezes at 273 K273\text{ K} and 1 bar1\text{ bar}. Adding 34.5 g34.5\text{ g} of ethanol into 500 g500\text{ g} of water alters the freezing point of the resultant solution. Take the molal freezing point depression constant of water to be 2 K kg mol−12\text{ K kg mol}^{-1}. Plots of vapour pressure (V.P.) against temperature (T) are shown below. [Molar mass of ethanol = 46 g mol−146\text{ g mol}^{-1}] Select the plot that correctly depicts the shift in freezing point.

Options

A
Option A
B
Option B
C
Option C
Correct
D
Option D

Topics & Concepts

SolutionsConcentration TermsAlcohols and Ethers

Step-by-Step Solution

To determine the correct plot depicting the shift in the freezing point, we perform the following step-by-step calculations and thermodynamic analysis:

1. Calculation of the Freezing Point of the Solution

Step A: Calculate the moles of ethanol (solute) Moles of ethanol (n)=Mass of ethanolMolar mass of ethanol=34.5 g46 g mol−1=0.75 mol\text{Moles of ethanol } (n) = \frac{\text{Mass of ethanol}}{\text{Molar mass of ethanol}} = \frac{34.5\text{ g}}{46\text{ g mol}^{-1}} = 0.75\text{ mol}

Step B: Calculate the molality of the solution (mm) Mass of solvent (water)=500 g=0.5 kg\text{Mass of solvent (water)} = 500\text{ g} = 0.5\text{ kg} m=Moles of ethanolMass of solvent in kg=0.75 mol0.5 kg=1.5 mol kg−1m = \frac{\text{Moles of ethanol}}{\text{Mass of solvent in kg}} = \frac{0.75\text{ mol}}{0.5\text{ kg}} = 1.5\text{ mol kg}^{-1}

Step C: Calculate the depression in freezing point (ΔTf\Delta T_f) ΔTf=Kf×m\Delta T_f = K_f \times m Given Kf=2 K kg mol−1K_f = 2\text{ K kg mol}^{-1}: ΔTf=2 K kg mol−1×1.5 mol kg−1=3 K\Delta T_f = 2\text{ K kg mol}^{-1} \times 1.5\text{ mol kg}^{-1} = 3\text{ K}

Step D: Calculate the new freezing point (TfT_f) Tf=Tf∘−ΔTf=273 K−3 K=270 KT_f = T_f^\circ - \Delta T_f = 273\text{ K} - 3\text{ K} = 270\text{ K}

This eliminates options [B] and [D], as they show a freezing point of 271 K271\text{ K}.

2. Analysis of the Vapour Pressure vs. Temperature Curves

  • The freezing point is the temperature at which the vapour pressure of the liquid phase equals the vapour pressure of the solid phase (ice).
  • According to the Clausius-Clapeyron equation, the vapour pressure of both solid and liquid states increases monotonically with an increase in temperature (dPdT>0\frac{dP}{dT} > 0). Therefore, the slope of the vapour pressure curves for pure water, ice, and the solution must be positive.
  • Option [A] incorrectly shows the vapour pressure of liquid water decreasing as temperature increases (negative slope).
  • Option [C] correctly shows:
    1. Positive slopes for all vapour pressure curves (Ice, Pure Water, and Water + Ethanol solution).
    2. The vapour pressure curve of the solution lying below that of pure water (lowering of vapour pressure).
    3. The intersection of the ice curve with the solution curve at 270 K270\text{ K} (the depressed freezing point) and with the pure water curve at 273 K273\text{ K}.

Thus, the plot in Option C correctly depicts the shift in freezing point.

Correct Option: C

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