Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes t

Q: Why is ethanol considered non-volatile here?

While ethanol is volatile in reality, in the context of freezing point depression problems where it's the solute in water, we treat it as a non-volatile solute that lowers the vapour pressure of the solvent, as implied by the options and standard problem-solving conventions.

Q: Why isn't the vapour pressure 1 bar at the freezing point?

A liquid boils when its vapour pressure equals the external pressure (1 bar). However, it freezes when its vapour pressure equals the vapour pressure of its solid phase. For water at 273 K, this equilibrium vapour pressure is very small (about 0.006 bar).

Q: How do we know the shape of the curves?

The vapour pressure of both solid and liquid phases increases exponentially with temperature (Clausius-Clapeyron equation). The solid (ice) curve is always steeper than the liquid (water) curve near the freezing point because the enthalpy of sublimation is greater than the enthalpy of vaporization.

Q: What does the addition of solute do to the graph?

Adding a solute lowers the vapour pressure of the liquid solvent at all temperatures. This shifts the liquid curve downwards. Consequently, it intersects the steep ice curve at a lower temperature, which is the visual representation of freezing point depression.

Q: Why did we use 2 for $K_f$ instead of the standard 1.86?

Always use the values provided in the question! The problem explicitly states to use $K_f = 2 \text{ K kg mol}^{-1}$ to simplify calculations. Using 1.86 would give a different $\Delta T_f$ and lead to confusion.

Question

Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as $2 	ext{ K kg mol}^{-1}$.

The figures shown below represents plots of vapour pressure (V.P.) versus temperature (T).

[Molecular weight of ethanol is $46 	ext{ g mol}^{-1}$]

Among the following, the option representing change in the freezing point is -

Sigma Percentile · Accepted Answer

This problem is a beautiful blend of mathematical calculation and conceptual graph analysis. It tests not just your ability to plug numbers into a formula, but your deep understanding of what a phase diagram actually represents.

### Decoding the Graph

Before we touch any numbers, let's understand the physical reality depicted in the graphs. The curves represent the vapour pressure of different phases as temperature changes. The steep curve is for solid ice, and the gentler curve is for liquid water.

**The freezing point is defined as the exact temperature where the vapour pressure of the liquid phase equals the vapour pressure of the solid phase.** Geometrically, this is the intersection point of the two curves.

### The Math of Depression

When we add a non-volatile solute (like ethanol in this specific context) to a solvent, it lowers the vapour pressure of the liquid at all temperatures. This shifts the entire liquid curve downwards. Because the liquid curve is now lower, it intersects the steep ice curve at a lower temperature. This is the visual proof of **freezing point depression**.

Let's calculate exactly how much the temperature drops. First, we find the molality ($m$) of the solution:

$$m = \frac{w_{	ext{solute}}}{M_{	ext{solute}} 	imes W_{	ext{solvent}} 	ext{ (in kg)}}$$

$$m = \frac{34.5}{46 	imes 0.5} = 1.5 	ext{ mol/kg}$$

Now, we apply the freezing point depression formula. Note that the problem explicitly asks us to use $K_f = 2 	ext{ K kg mol}^{-1}$ to keep the math clean:

$$\Delta T_f = K_f 	imes m = 2 	imes 1.5 = 3 	ext{ K}$$

Since pure water freezes at $273 	ext{ K}$, our new freezing point is:

$$T_f = 273 - 3 = 270 	ext{ K}$$

This immediately eliminates options (A) and (B), which show an intersection at $271 	ext{ K}$.

### The One Bar Trap

Now we are left with options (C) and (D). Both show the correct freezing point of $270 	ext{ K}$. So, what is the difference? Look closely at the y-axis.

In graph (C), the intersection of the curves happens exactly at the $1 	ext{ bar}$ mark. In graph (D), the intersection happens far below the $1 	ext{ bar}$ mark.

This is where many students fall into a trap. We know water freezes at $1 	ext{ bar}$ of **external atmospheric pressure**. However, the y-axis of this graph is **vapour pressure**, not external pressure!

At its freezing point ($273 	ext{ K}$), the vapour pressure of water is extremely low—approximately $0.006 	ext{ bar}$. It only reaches $1 	ext{ bar}$ at its normal boiling point ($373 	ext{ K}$). Therefore, the intersection of the ice and water curves must occur at a vapour pressure significantly lower than $1 	ext{ bar}$.

### Final Conclusion

Graph (D) perfectly captures both realities: the new freezing point is correctly positioned at $270 	ext{ K}$, and the intersection points correctly lie well below the $1 	ext{ bar}$ vapour pressure line. This makes it the only scientifically accurate representation of the system.

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Visualized Solution

$Understanding the Goal$

$Calculating Molality$

$Δ T_{f} = K_{f} \times m$

$T_{f} = T_{f}^{\circ} - Δ T_{f}$

$Analyzing the Vapour Pressure$

$Selecting the Correct Graph$

The Sigma Insight: Colligative Properties

Decoding the Graph

The Math of Depression

The One Bar Trap

Final Conclusion

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Understanding the Goal

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Molality

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔTf​=Kf​×m

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tf​=Tf∘​−ΔTf​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Vapour Pressure

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Selecting the Correct Graph