Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be

Q: Why do we use molality instead of molarity in this formula?

Molality is temperature-independent because it is based on mass, whereas molarity depends on volume, which changes with temperature. Since freezing involves temperature changes, molality is the accurate measure.

Q: What exactly does $K_f$ represent?

$K_f$ is the molal depression constant (or cryoscopic constant). It represents the depression in freezing point when 1 mole of a non-volatile solute is dissolved in 1 kg of solvent.

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

Solute particles disrupt the formation of the orderly crystal lattice of the solid solvent. The system must be cooled to a lower temperature to achieve the necessary decrease in kinetic energy for freezing to occur.

Q: Can we use any solute as an antifreeze?

While any non-volatile solute will lower the freezing point, practical antifreezes like ethylene glycol are chosen because they are highly soluble in water, non-corrosive, and have a high boiling point (acting as a coolant in summer too).

Q: Why did we convert 4 kg to 4000 g?

The standard formula for molality often uses the mass of the solvent in kg. If you use the formula $m = \frac{w_1 \times 1000}{M_1 \times w_2}$, $w_2$ must be in grams. Alternatively, you can keep it as 4 kg and omit the 1000 in the numerator. Both yield the same result!

Question

Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to $4	ext{ kg}$ of water to prevent it from freezing at $-6^\circ	ext{C}$ will be ($K_f$ for water $=1.86	ext{ K kg mol}^{-1}$ and molar mass of ethylene glycol $=62	ext{ g mol}^{-1}$)

Options:

(A) $804.32	ext{ g}$

(B) $204.30	ext{ g}$

(C) $400.00	ext{ g}$

(D) $304.60	ext{ g}$

Sigma Percentile · Accepted Answer

### The Freezing Challenge

Imagine you are driving through a freezing landscape. The water in your car's radiator is at risk of turning into solid ice, which could expand and destroy the engine. To prevent this, we need to lower the freezing point of the water.

This is where the magic of **Colligative Properties** comes into play. By adding a non-volatile solute like ethylene glycol, we can disrupt the water molecules' ability to form a solid crystal lattice.

### Decoding the Depression in Freezing Point

The problem states that we need to protect $4	ext{ kg}$ of water down to a temperature of $-6^\circ	ext{C}$. Pure water freezes at $0^\circ	ext{C}$.

Therefore, the required depression in freezing point, denoted by $\Delta T_f$, is exactly $6	ext{ K}$ (or $6^\circ	ext{C}$).

The fundamental law governing this phenomenon is:
$$ \Delta T_f = K_f 	imes m $$

Here, $K_f$ is the **molal depression constant** (given as $1.86	ext{ K kg mol}^{-1}$), and $m$ is the **molality** of the solution.

### Setting Up the Master Equation

Molality is defined as the number of moles of solute per kilogram of solvent. Let the mass of ethylene glycol we need to add be $w_1$.

The molar mass of ethylene glycol ($C_2H_6O_2$) is $62	ext{ g mol}^{-1}$.

We can expand the molality term:
$$ m = \frac{w_1 	imes 1000}{M_1 	imes w_2} $$

Substituting this back into our main equation gives:
$$ \Delta T_f = \frac{K_f 	imes w_1 	imes 1000}{M_1 	imes w_2} $$

### The Calculation Anomaly

Now, let's substitute our known values into the equation:
$$ 6 = \frac{1.86 	imes w_1 	imes 1000}{62 	imes 4000} $$

We can simplify the thousands immediately, leaving a $4$ in the denominator:
$$ 6 = \frac{1.86 	imes w_1}{62 	imes 4} $$

Rearranging to solve for $w_1$:
$$ w_1 = \frac{6 	imes 62 	imes 4}{1.86} $$
$$ w_1 = \frac{1488}{1.86} $$

If you perform this exact division, you get **$800	ext{ g}$**.

However, looking at the given options, $800	ext{ g}$ is missing! The closest option is **$804.32	ext{ g}$**. This slight discrepancy occurs because the question setter likely used a slightly different value for $K_f$ (specifically, $1.85	ext{ K kg mol}^{-1}$) during their internal calculations.

In competitive exams like JEE, when faced with such an anomaly, always trust your rigorous calculation and select the closest matching option. Thus, we confidently choose option (a).

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Visualized Solution

$The Setup$

$Depression in Freezing Point$

$The Formula$

$Substitution$

$Calculation$

$Final Answer$

$The Way Forward$

The Sigma Insight: Colligative Properties

The Freezing Challenge

Decoding the Depression in Freezing Point

Setting Up the Master Equation

The Calculation Anomaly

Similar Questions

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40 g of glucose (Molar mass = 180) is mixed with 200 mL of water. The freezing point of solution is ............ K. (Nearest integer) [Given, $K_{f} = 1.86 K kg mol^{- 1}$ , density of water $= 1.00 g cm^{- 3}$ , freezing point of water $= 273.15 K$ ]

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Depression in Freezing Point

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

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

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

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

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