An antifreeze solution is prepared by dissolving 31 g of ethylene glycol (Molar mass $= 62$ g mol$^{-1}$) in 600 g of water. Calculate the freezing point of the solution. ($K_f$ for water $= 1.86$ K kg mol$^{-1}$)

An antifreeze solution is prepared by dissolving 31 g of ethylene glycol (Molar mass = 62 g mol⁻¹) in 600 g of

Question

= 62

CBSECBSE Class XII Board 2026Subjective· 3mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Mass Percentage

Mass Percentage: The Intuition

Imagine you're making lemonade. You mix 50 grams of sugar into 200 grams of water. The total drink weighs 250 grams. Now, if someone asks, "How much of this drink is actually sugar?" — you're not just saying "50 grams." You want to say what fraction of the whole mixture is sugar, scaled to a convenient 100.

That's mass percentage. It answers: "Out of every 100 grams of the mixture, how many grams are this particular component?"

In our lemonade, sugar is 50 g out of 250 g total. That's $\frac{50}{250} = 0.2$ of the whole. Multiply by 100 to get the percentage: $0.2 \times 100 = 20\%$ . So, 20% of the drink's mass is sugar. If you had 100 g of this lemonade, 20 g of it would be sugar.

The Precise Definition

\text{Mass percentage of component} = \frac{\text{Mass of that component}}{\text{Total mass of mixture}} \times 100\%

The formula is simple, but the key is understanding what "total mass" means. It's the sum of masses of all components in the mixture — nothing more, nothing less.

Why It Matters in Chemistry

Mass percentage is one of the most common ways to express concentration — how much of a substance is present in a mixture. You'll see it in:

Solutions: "10% salt water" means 10 g of salt dissolved in enough water to make 100 g of solution (not 10 g salt + 100 g water — that would be 110 g total, giving only about 9.1%).
Alloys: "18-karat gold" is 75% gold by mass (18 parts gold out of 24 total parts).
Food labels: "Fat: 15%" means 15 g of fat per 100 g of the food.

Watch out

A Common Mistake

Students often think "10% salt solution" means 10 g salt + 100 g water. That's wrong. It means 10 g salt + 90 g water = 100 g total solution. The denominator is total mass, not the mass of the solvent alone.

Step-by-Step Example

Problem: A solution is made by dissolving 25 g of glucose in 175 g of water. Find the mass percentage of glucose.

Step 1: Identify the component you care about — glucose (25 g).

Step 2: Find the total mass of the mixture.

\text{Total mass} = 25\ \text{g (glucose)} + 175\ \text{g (water)} = 200\ \text{g}

Step 3: Apply the formula.

$\text{Mass percentage of glucose} = \frac{25}{200} \times 100\% = 12.5\%$ …

Why this formula?

Let's break down Mass Percentage from first principles. The goal is to understand why the formula is what it is, not just to memorize it.

1. The Core Idea: "Part of a Whole"

Mass percentage answers a simple question: "If I break a mixture into 100 equal parts by mass, how many of those parts come from a specific component?"

Imagine you have a bowl of fruit salad. The total mass is 500 grams. The apples in it weigh 100 grams.

The apples are a part of the whole salad.
The whole salad is the total.

The mass percentage tells you the fraction of the total mass that is apples, but expressed "out of 100" (per cent).

2. The Natural First Step: The Fraction

Before we talk about "percentage," we talk about the fraction of the total:

\text{Fraction of component} = \frac{\text{Mass of component}}{\text{Total mass of mixture}}

For the apple example:

\frac{100 \text{ g}}{500 \text{ g}} = 0.2

This means 0.2 (or one-fifth) of the total mass is apples. This is the pure ratio — no scaling yet.

3. Why Multiply by 100?

A fraction like 0.2 is perfectly correct, but it's not intuitive for quick comparison. "Per cent" literally means "per hundred" (from Latin per centum).

To convert a fraction into a "per hundred" number, we multiply by 100:

\text{Percentage} = (\text{Fraction}) \times 100

So:

0.2 \times 100 = 20\%

This tells us: "Out of every 100 grams of fruit salad, 20 grams come from apples." That's much easier to visualize.

4. The Final Formula (The "Why" in One Line)

Putting the fraction and the "times 100" together gives the standard formula:

\text{Mass percentage} = \frac{\text{Mass of component}}{\text{Total mass of mixture}} \times 100\%

Why does this work?

Because it's just:

Find the proportion (part ÷ whole).
Scale that proportion to per hundred (× 100).

5. A Common Exam Trap (and Why It's Wrong)

Sometimes students write: …

Start your 14-day free trial to unlock the full solution →

The freezing point of a solution is lowered by an amount proportional to the molality of the solute. Here, dissolving 31 g of ethylene glycol in 600 g of water gives a molality of 0.833 mol/kg, leading to a freezing point depression of 1.55 K. The solution freezes at –1.55 °C.

When you add a non-volatile solute like ethylene glycol to water, the freezing point of the water drops. This is a colligative property — it depends only on how many solute particles are present, not on what they are. The key formula is:

\Delta T_f = K_f \cdot m

where $\Delta T_f$ is the freezing point depression, $K_f$ is the cryoscopic constant (1.86 K kg mol⁻¹ for water), and $m$ is the molality of the solution.

Molality is moles of solute per kilogram of solvent. That’s the natural unit here because it doesn’t change with temperature, unlike molarity.

Let’s work through it.

Find moles of ethylene glycol Mass given = 31 g, molar mass = 62 g mol⁻¹.

\text{Moles} = \frac{31}{62} = 0.5 \text{ mol}

Find mass of solvent in kilograms

Water mass = 600 g = 0.600 kg.
Calculate molality

m = \frac{0.5 \text{ mol}}{0.600 \text{ kg}} = 0.8333 \text{ mol/kg}

Apply the freezing point depression formula $\Delta T_f = 1.86 \times 0.8333 = 1.55 \text{ K}$ …

Start your 14-day free trial to unlock the full solution →

Q.An antifreeze solution is prepared by dissolving 31 g of ethylene glycol (Molar mass =62= 62=62 g mol−1^{-1}−1) in 600 g of water. Calculate the freezing point of the solution. (KfK_fKf​ for water =1.86= 1.86=1.86 K kg mol−1^{-1}−1)

🎓Unlock everything free for 14 days

Q.An antifreeze solution is prepared by dissolving 31 g of ethylene glycol (Molar mass $= 62$ g mol $^{-1}$ ) in 600 g of water. Calculate the freezing point of the solution. ( $K_f$ for water $= 1.86$ K kg mol $^{-1}$ )

Unlock everything free for 14 days