83 g of ethylene glycol dissolved in 625 g of water. The freezing point of the solution is ………K. (Ne

Q: Why do we use molality instead of molarity for freezing point depression?

Molality is temperature-independent because it is based on the mass of the solvent. Molarity depends on the volume of the solution, which changes with temperature. Since freezing involves a temperature change, molality is the reliable choice!

Q: Why is the van't Hoff factor ($i$) equal to 1 for ethylene glycol?

Ethylene glycol ($C_2H_6O_2$) is a covalent compound and a non-electrolyte. When it dissolves in water, it stays as intact molecules rather than breaking apart into ions. Therefore, 1 molecule yields 1 particle in solution, making $i = 1$.

Q: What would happen if we used $83 \text{ g}$ of NaCl instead of ethylene glycol?

NaCl is a strong electrolyte that dissociates into $Na^+$ and $Cl^-$ ions, giving a van't Hoff factor of $i = 2$. This would roughly double the freezing point depression, making the solution freeze at an even lower temperature!

Q: Why does adding a solute lower the freezing point?

Solute particles disrupt the formation of the orderly solid lattice of the solvent. The solvent molecules need even less kinetic energy (a lower temperature) to overcome this disruption and lock into a solid state.

Q: Can the freezing point depression be negative?

The term "depression" ($\Delta T_f$) itself represents a positive drop in temperature. However, the final freezing point ($T_f$) can be negative on the Celsius scale, but it will always be a positive value on the Kelvin scale.

Question

83 g of ethylene glycol dissolved in 625 g of water. The freezing point of the solution is ………K. (Nearest integer)
[Use, molal freezing point depression constant of water $= 1.86 	ext{ K kg mol}^{-1}$, Freezing point of water $= 273 	ext{ K}$ and Atomic masses : $	ext{C} = 12.0 	ext{ u}, 	ext{O} = 16.0 	ext{ u}, 	ext{H} = 1.0 	ext{ u}$]

Sigma Percentile · Accepted Answer

### The Antifreeze Phenomenon

Have you ever wondered how car radiators survive freezing winters without the water inside turning into solid ice? The secret lies in a fascinating concept called **freezing point depression**.

When we add a non-volatile solute to a pure solvent, the freezing point of the resulting solution drops. In this problem, we are acting as chemical engineers, adding ethylene glycol (a common antifreeze) to water.

### Decoding the Solute

Our first task is to understand our solute, ethylene glycol. Its chemical formula is $C_2H_6O_2$.

To find its molar mass, we sum the atomic masses of its constituent atoms:

$$M = (2 	imes 12) + (6 	imes 1) + (2 	imes 16) = 62 	ext{ g mol}^{-1}$$

This tells us that every mole of ethylene glycol weighs exactly $62 	ext{ g}$.

### Concentration in Molality

Colligative properties like freezing point depression depend on the number of solute particles, not their identity. The best way to express this concentration is through **molality** ($m$).

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

$$m = \frac{	ext{Mass of solute} / 	ext{Molar mass}}{	ext{Mass of solvent in kg}}$$

Substituting our values:

$$m = \frac{83 / 62}{625 / 1000} = \frac{83 	imes 1000}{62 	imes 625}$$

### The Master Equation

Now, we bring in the master equation for freezing point depression:

$$\Delta T_f = i \cdot K_f \cdot m$$

Here, $i$ is the **van't Hoff factor**. Since ethylene glycol is a non-electrolyte and does not dissociate into ions in water, $i = 1$.

The constant $K_f$ is the molal depression constant for water, given as $1.86 	ext{ K kg mol}^{-1}$.

Plugging everything in:

$$\Delta T_f = 1 	imes 1.86 	imes \frac{83 	imes 1000}{62 	imes 625}$$

### The Final Chill

Let's crunch the numbers. Simplifying the fraction gives us:

$$\Delta T_f = 1.86 	imes 2.1419 \approx 3.98 	ext{ K}$$

This means the freezing point of our water has dropped by $3.98 	ext{ K}$.

To find the final freezing point of the solution ($T_f$), we subtract this depression from the normal freezing point of pure water ($273 	ext{ K}$):

$$T_f = 273 - 3.98 = 269.02 	ext{ K}$$

Rounding to the nearest integer, we get our final answer: **$269$**.

Enter Numerical Value:

Visualized Solution

$M o l a r M a ss o f E t h y l e n e Gl y co l$

$M o l a l i t y o f t h e S o l u t i o n$

$D e p r ess i o n in F r eez in g P o in t$

$C a l c u l a t in g Δ T_{f}$

$F ina l F r eez in g P o in t$

The Sigma Insight: Colligative Properties

The Antifreeze Phenomenon

Decoding the Solute

Concentration in Molality

The Master Equation

The Final Chill

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