A solution contain 62 g of ethylene glycol in 250 g of water is cooled upto -10°C. If K_f for water

Q: Why doesn't the ethylene glycol freeze along with the water?

Ethylene glycol acts as a solute that lowers the freezing point of the entire mixture. When the temperature drops, the pure solvent (water) begins to freeze out first, leaving behind a more concentrated solution of ethylene glycol.

Q: Why is $\Delta T_f$ taken as 10 and not -10?

$\Delta T_f$ represents the *depression* or the *drop* in the freezing point. It is calculated as $T_f^\circ - T_f$, which is $0 - (-10) = 10\text{ K}$. It is always a positive quantity.

Q: What exactly does $w_A$ represent in the molality formula here?

This is a common trap! $w_A$ is the mass of the solvent that is *currently* dissolving the solute. Since the ice has separated out, it is no longer acting as a solvent. So, $w_A$ is only the mass of the liquid water.

Q: How did we get 62 as the molar mass of ethylene glycol?

The chemical formula for ethylene glycol is $C_2H_6O_2$. Calculating its molar mass: $(2 \times 12) + (6 \times 1) + (2 \times 16) = 24 + 6 + 32 = 62\text{ g/mol}$.

Q: Will all the water eventually freeze if we keep cooling it?

If you keep lowering the temperature, more and more water will freeze, making the remaining solution even more concentrated. Eventually, you will reach the eutectic point where the entire mixture (solute and solvent) freezes solid.

Question

A solution contain $62	ext{ g}$ of ethylene glycol in $250	ext{ g}$ of water is cooled upto $-10^\circ	ext{C}$. If $K_f$ for water is $1.86	ext{ K kg mol}^{-1}$, then amount of water (in g) separated as ice is

Options:

(A) 32

(B) 48

(C) 64

(D) 16

Sigma Percentile · Accepted Answer

### The Chilling Setup

Imagine you are in a laboratory, holding a beaker filled with a mixture of $250	ext{ g}$ of pure water and $62	ext{ g}$ of ethylene glycol ($C_2H_6O_2$). You place this beaker into a deep freezer and watch the temperature drop steadily. As the temperature plunges below $0^\circ	ext{C}$, something fascinating happens. The water doesn't freeze all at once. Instead, because of the dissolved ethylene glycol, the freezing point of the solution is lowered.

However, as the temperature reaches $-10^\circ	ext{C}$, the solution can no longer keep all the water in the liquid phase. Pure water begins to crystallize and separate out as solid ice. Our mission is to find out exactly how much of that original $250	ext{ g}$ of water has turned into ice.

### The Hidden Trap: What is the Solvent?

To solve this, we need to use the concept of **Depression in Freezing Point**. The formula is elegantly simple:

$$ \Delta T_f = K_f 	imes m $$

Where $m$ is the molality of the solution. Expanding the molality term, we get:

$$ \Delta T_f = K_f 	imes \frac{w_B}{M_B} 	imes \frac{1000}{w_A} $$

Here lies the biggest trap in this problem! What exactly is $w_A$? Many students mistakenly plug in the initial $250	ext{ g}$ of water. But remember, molality is defined by the mass of the solvent that is *currently* dissolving the solute. The water that has turned into ice is no longer acting as a solvent; it has separated from the solution. Therefore, $w_A$ represents **only the mass of the water that remains in the liquid state** at $-10^\circ	ext{C}$.

### The Master Equation: Depression in Freezing Point

Let's gather our known variables and substitute them into our master equation. The freezing point of pure water is $0^\circ	ext{C}$, and our final temperature is $-10^\circ	ext{C}$. The depression in freezing point ($\Delta T_f$) is always a positive drop:

$$ \Delta T_f = 0^\circ	ext{C} - (-10^\circ	ext{C}) = 10	ext{ K} $$

We are given the cryoscopic constant for water, $K_f = 1.86	ext{ K kg mol}^{-1}$. The mass of our solute, ethylene glycol ($w_B$), is $62	ext{ g}$. We also need its molar mass ($M_B$). The chemical formula is $C_2H_6O_2$, so:

$$ M_B = (2 	imes 12) + (6 	imes 1) + (2 	imes 16) = 62	ext{ g/mol} $$

Now, let's plug everything into the equation:

$$ 10 = 1.86 	imes \frac{62}{62} 	imes \frac{1000}{w_A} $$

Notice the beautiful mathematical elegance here! The $62$ in the numerator perfectly cancels out the $62$ in the denominator. The equation simplifies dramatically:

$$ 10 = 1.86 	imes 1 	imes \frac{1000}{w_A} $$

$$ 10 = \frac{1860}{w_A} $$

Solving for $w_A$, we get:

$$ w_A = \frac{1860}{10} = 186	ext{ g} $$

This tells us that at $-10^\circ	ext{C}$, exactly $186	ext{ g}$ of water is still in the liquid phase, actively dissolving the ethylene glycol.

### The Final Calculation: Finding the Ice

We are almost at the finish line. The question asks for the amount of water that has separated as ice. We started with an initial water mass of $250	ext{ g}$. We just calculated that $186	ext{ g}$ of it is still liquid. The difference between these two values must be the mass of the ice that formed.

$$ 	ext{Mass of Ice} = W_{	ext{initial}} - w_A $$

$$ 	ext{Mass of Ice} = 250	ext{ g} - 186	ext{ g} = 64	ext{ g} $$

And there we have it! Exactly **$64	ext{ g}$** of ice has separated from the solution. This principle isn't just a textbook problem; it's the exact science behind how antifreeze works in car radiators during harsh winters, preventing the engine block from cracking due to expanding ice!

Animated Solution for Chemistry - Solutions: A solution contain $62 g$ of ethylene glycol in $250 g$ of water is cooled upto $- 1 0^{\circ} C$ . If $K_{f}$ for water is $1.86 K kg mol^{- 1}$ , then amount of water (in g) separated as ice is

Select Answer:

Visualized Solution

\text{Visualizing the Cooling Process}

\text{Depression in Freezing Point}

\text{Substituting the Values}

\text{Calculating Liquid Water Mass}

\text{Calculating Mass of Ice}

\text{The Way Forward}

The Sigma Insight: Colligative Properties

The Chilling Setup

The Hidden Trap

The Master Equation

The Final Calculation

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