A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL$^{-1}$, then what shall be the molarity of the solution?

A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each co

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Concept understanding — Molality Calculation

Molality: The Concentration That Ignores Temperature

Imagine you're making a cup of sweet tea. You add sugar to hot water, stir, and taste. If you let the tea cool to room temperature, the amount of sugar hasn't changed — but the volume of the liquid has shrunk slightly. If you measured concentration as "grams of sugar per litre of solution," that number would change just because the temperature changed. That's annoying if you're a chemist who needs a reliable, temperature-independent way to describe how much solute is present.

Molality was invented to solve exactly this problem.

The Intuition

Instead of measuring the volume of the solution (which expands and contracts with temperature), molality measures the mass of the solvent. Mass doesn't change with temperature. So molality gives you a concentration that stays the same whether your solution is hot or cold.

Think of it this way:

Molarity = moles of solute per litre of solution (temperature-sensitive)
Molality = moles of solute per kilogram of solvent (temperature-independent)

The solvent is the substance doing the dissolving — usually water. The solute is what gets dissolved — sugar, salt, etc.

The Precise Definition

\text{Molality } (m) = \frac{\text{moles of solute}}{\text{kilograms of solvent}}

The symbol for molality is a lowercase m (not to be confused with M for molarity).

Key points to remember:

The denominator is solvent mass, not solution mass
The unit is mol/kg (often written as simply "m")
It is independent of temperature because mass doesn't change with temperature

Worked Example

Problem: 36 g of glucose ( $C_6H_{12}O_6$ , molar mass = 180 g/mol) is dissolved in 500 g of water. Calculate the molality of the solution.

Step 1: Find moles of solute

\text{Moles of glucose} = \frac{36 \text{ g}}{180 \text{ g/mol}} = 0.2 \text{ mol}

Step 2: Convert solvent mass to kilograms

500 \text{ g} = 0.5 \text{ kg}

Step 3: Apply the formula

m = \frac{0.2 \text{ mol}}{0.5 \text{ kg}} = 0.4 \text{ m}

Note

The answer is 0.4 m (or 0.4 mol/kg). Notice we used the mass of water (500 g), not the mass of the solution (which would be 536 g).

Common Mistake to Avoid

Watch out

Do not use the mass of the solution in the denominator. The formula specifically asks for the mass of the solvent alone. If the problem gives you the total mass of the solution, subtract the mass of the solute to find the solvent mass.

When Do You Use Molality?

Molality is the star in two important situations: …

Why this formula?

Molality Calculation: Why the Formula Works

Molality is a measure of concentration that is temperature-independent — this is its key advantage over molarity. Let's understand why the formula takes the form it does.

The Definition First

Molality ( $m$ ) is defined as:

m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}

The unit is mol/kg, often written as m (e.g., 0.5 m glucose solution).

Why Mass of Solvent, Not Solution?

This is the critical conceptual point.

The Reasoning

Molarity uses volume of solution → volume changes with temperature (expansion/contraction). So molarity changes with temperature.
Molality uses mass of solvent → mass is invariant with temperature. So molality remains constant regardless of temperature changes.

Key insight: By using the solvent's mass (not the solution's volume), we eliminate temperature dependence. This is why molality is preferred for colligative properties (boiling point elevation, freezing point depression) — these properties depend on the number of solute particles, not on temperature.

Deriving the Formula Step-by-Step

Step 1: Moles of Solute

If you have $w_{\text{solute}}$ grams of solute with molar mass $M_{\text{solute}}$ (g/mol):

\text{moles of solute} = \frac{w_{\text{solute}}}{M_{\text{solute}}}

Step 2: Mass of Solvent in kg

If the solvent mass is $W_{\text{solvent}}$ grams:

\text{mass of solvent in kg} = \frac{W_{\text{solvent}}}{1000}

Step 3: Putting It Together

m = \frac{\frac{w_{\text{solute}}}{M_{\text{solute}}}}{\frac{W_{\text{solvent}}}{1000}}

Simplifying:

m = \frac{w_{\text{solute}} \times 1000}{M_{\text{solute}} \times W_{\text{solvent}}}

The Final Formula (Exam-Ready)

\boxed{m = \frac{w_{\text{solute}} \times 1000}{M_{\text{solute}} \times W_{\text{solvent}}}}

Where:

$w_{\text{solute}}$ = mass of solute in grams
$M_{\text{solute}}$ = molar mass of solute in g/mol
$W_{\text{solvent}}$ = mass of solvent in grams

Why the ×1000 Factor? …

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The key idea is to interpret 10% w/w as 10 g glucose per 100 g solution, then use the definitions of molality (moles of solute per kg of solvent), mole fraction, and molarity (moles per litre of solution, using density). The molality is 0.617 m, the mole fraction of glucose is 0.011, of water is 0.989, and the molarity is 0.667 M.

Let’s unpack this step by step. The problem gives a “10% w/w” glucose solution — that means 10 grams of glucose are present in every 100 grams of the solution. The rest (90 g) is water, the solvent. This is the starting point for all three quantities.

1. Molality — moles of solute per kg of solvent

Molality ( $m$ ) is defined as:

m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}

First, find moles of glucose. Glucose is $\text{C}_6\text{H}_{12}\text{O}_6$ , molar mass = $6 \times 12 + 12 \times 1 + 6 \times 16 = 72 + 12 + 96 = 180 \text{ g mol}^{-1}$ .

In 100 g of solution, we have 10 g glucose. So:

\text{moles of glucose} = \frac{10}{180} = \frac{1}{18} \approx 0.05556 \text{ mol}

Mass of solvent (water) = $100 - 10 = 90 \text{ g} = 0.090 \text{ kg}$ .

Thus:

m = \frac{0.05556}{0.090} = 0.6173 \text{ mol kg}^{-1}

Tip

Notice that molality depends only on the ratio of solute to solvent mass — it is independent of temperature and density. That’s why it’s preferred for colligative properties.

2. Mole fraction of each component

Mole fraction ( $x$ ) is moles of one component divided by total moles in the solution.

We already have moles of glucose = $0.05556$ .

Moles of water: mass of water = 90 g, molar mass = 18 g mol $^{-1}$ .

\text{moles of water} = \frac{90}{18} = 5.00 \text{ mol}

Total moles = $0.05556 + 5.00 = 5.05556$ mol.

Mole fraction of glucose:

x_{\text{glucose}} = \frac{0.05556}{5.05556} \approx 0.0110

Mole fraction of water:

x_{\text{water}} = \frac{5.00}{5.05556} \approx 0.9890

Check: $0.0110 + 0.9890 = 1.0000$ — good. …

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🧠 Common Mistake #1: Confusing % w/w with % w/v or % v/v

The error:

Students treat “10% w/w” as 10 g of glucose in 100 mL of solution (which is % w/v) or 10 g in 100 g of solution (correct for w/w) but then incorrectly use volume for molality.

How to avoid:

% w/w = mass of solute per 100 g of solution (not per 100 mL).
Always write:

10% w/w → 10 g glucose + 90 g water (total 100 g solution).
Molality uses mass of solvent in kg, not mass of solution.

🧠 Common Mistake #2: Using density at the wrong step

The error:

Students plug density into molality calculation. Molality does not involve volume or density — only masses.

How to avoid:

Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}}$
Density is needed only for molarity (which uses volume of solution).
Keep them separate:
- Molality → use masses.
- Molarity → use density to get volume from mass of solution.

🧠 Common Mistake #3: Forgetting to convert grams to kg for molality

The error:

Using mass of solvent in grams directly in the molality formula.

How to avoid:

Always convert:

$90 \, \text{g} = 0.090 \, \text{kg}$
Formula:

m = \frac{n_{\text{solute}}}{m_{\text{solvent (kg)}}}

🧠 Common Mistake #4: Wrong mole fraction formula

The error:

Using mass instead of moles, or forgetting that mole fraction of all components must sum to 1.

How to avoid:

Mole fraction of glucose:

x_{\text{glucose}} = \frac{n_{\text{glucose}}}{n_{\text{glucose}} + n_{\text{water}}}

Mole fraction of water:

x_{\text{water}} = 1 - x_{\text{glucose}}

Always verify: $x_{\text{glucose}} + x_{\text{water}} = 1$

🧠 Common Mistake #5: Using density of pure water instead of solution density for molarity

The error:

Assuming density of solution = 1 g/mL (like pure water) when it’s given as 1.2 g/mL.

How to avoid:

Read the problem carefully — density is of the solution, not solvent.
For molarity:
- Mass of 100 g solution → Volume = $\frac{100 \, \text{g}}{1.2 \, \text{g/mL}} = 83.33 \, \text{mL}$
- Convert to L: $0.08333 \, \text{L}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
$0.892\,N$
›Reveal solutionSolution

" $20$ volume" means $1\ \text{L}$ of solution releases $20\ \text{L}$ of O $_2$ at STP; using $N=\dfrac{\text{volume strength}}{5.6}$ gives $N=\dfrac{20}{5.6}=3.57\ N$ - option (D).

Meaning of volume strength. A " $20$ volume" H $_2$ O $_2$ solution liberates $20\ \text{L}$ of O $_2$ (at STP) per litre of solution on decomposition:
$2\,\text{H}_2\text{O}_2\rightarrow2\,\text{H}_2\text{O}+\text{O}_2$
Step 1 - Moles of O $_2$ per litre.
$n_{\text{O}_2}=\frac{20}{22.4}=0.893\ \text{mol}$
Step 2 - Moles and equivalents of H $_2$ O $_2$ . …
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TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
$_3$
›Reveal solutionSolution
The key is to use the balanced chemical equation and stoichiometry to find the mass of pure HCl needed, then convert to the mass of the 50% w/w solution. The required mass is 292 g, so option (B) is correct.

Concept & Intuition

This problem is about reacting hydrochloric acid with calcium carbonate. The reaction is a classic acid–carbonate neutralization:
$\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O}$
We are given a 50% w/w solution — meaning 50 g of pure HCl per 100 g of solution. So the actual mass of solution needed will be double the mass of pure HCl required. The trap is forgetting to account for the dilution and just picking the mass of pure HCl.

Step-by-step solution

Write the balanced equation

$\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O}$
This tells us: 1 mole of CaCO₃ reacts with 2 moles of HCl.

Find moles of CaCO₃ Molar mass of CaCO₃ = 40 (Ca) + 12 (C) + 3×16 (O) = 100 g/mol. Given mass = 200 g.

$\text{Moles of CaCO}_3 = \frac{200}{100} = 2 \text{ mol}$

Find moles of HCl needed

From the equation: 1 mol CaCO₃ needs 2 mol HCl.

So 2 mol CaCO₃ need $2 \times 2 = 4$ mol HCl.

Find mass of pure HCl required

Molar mass of HCl = 1 + 35.5 = 36.5 g/mol.

$\text{Mass of pure HCl} = 4 \times 36.5 = 146 \text{ g}$ …
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TG EAPCET 2021Set ap-2021-08-10-AN1 markMCQ
$\mathrm{H_2O_2}$
›Reveal solutionSolution
The "volume strength" of hydrogen peroxide is the volume of $\mathrm{O_2}$ (at STP) released per volume of solution. For a 50-volume solution the strength (w/v) is about 15%, so the correct option is (D).

Concept & Intuition

"Volume strength" (e.g. "10 volume," "50 volume") labels a hydrogen peroxide solution by the volume of oxygen it releases: 1 mL of the solution produces that many mL of $\mathrm{O_2}$ at STP on decomposition. The decomposition is
$2\mathrm{H_2O_2} \rightarrow 2\mathrm{H_2O} + \mathrm{O_2}$
To convert volume strength into a percentage (g of $\mathrm{H_2O_2}$ per 100 mL of solution), use the molar volume at STP (22.4 L/mol) and the molar mass of $\mathrm{H_2O_2}$ (34 g/mol).

Step-by-step reasoning

Interpret "50 volume"

1 mL of solution yields 50 mL of $\mathrm{O_2}$ at STP, so 1 L of solution yields $50 \times 1000 = 50\,000$ mL = 50 L of $\mathrm{O_2}$ .

Moles of $\mathrm{O_2}$

$n_{\mathrm{O_2}} = \frac{50}{22.4} \approx 2.232\ \text{mol}$

Moles of $\mathrm{H_2O_2}$

From the 2:1 ratio, $n_{\mathrm{H_2O_2}} = 2 \times 2.232 = 4.464$ mol.

Mass of $\mathrm{H_2O_2}$ per litre

With molar mass 34 g/mol,

$m = 4.464 \times 34 \approx 151.8\ \text{g per litre}$

Express as a percentage (w/v) …
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TG EAPCET 2021Set ap-2021-08-10-FN1 markMCQ
$\mathrm{H_2O_2}$
›Reveal solutionSolution
"Volume strength" of $\mathrm{H_2O_2}$ is the volume of $\mathrm{O_2}$ (at STP) released per volume of solution. For a 50-volume solution, converting through the decomposition $2\mathrm{H_2O_2} \to 2\mathrm{H_2O} + \mathrm{O_2}$ gives a strength of about 15%, so the correct option is (D).

Why this approach works

"Volume strength" tells you how many millilitres of oxygen gas (at STP) one millilitre of the solution releases on decomposition. So a "50 volume" solution means 1 mL of solution yields 50 mL of $\mathrm{O_2}$ .

To convert this into a percentage by mass (w/v), we:

find the mass of $\mathrm{H_2O_2}$ that produces that volume of oxygen, using the decomposition stoichiometry;

relate that mass to the mass of solution (density approx 1 g/mL for dilute solutions).

The bridge is the balanced equation:
$2\mathrm{H_2O_2} \rightarrow 2\mathrm{H_2O} + \mathrm{O_2}$
so 2 moles of $\mathrm{H_2O_2}$ give 1 mole of $\mathrm{O_2}$ .

Step-by-step reasoning

Moles of oxygen from the given volume At STP, 1 mole of gas occupies 22400 mL. For 50 mL of $\mathrm{O_2}$ (from 1 mL of solution):

$n_{\mathrm{O_2}} = \frac{50}{22400} = \frac{1}{448}\ \text{mol}$

Moles of $\mathrm{H_2O_2}$ that produce this oxygen

From the 2:1 ratio, $n_{\mathrm{H_2O_2}} = 2 \times \frac{1}{448} = \frac{1}{224}\ \text{mol}$ .

Mass of $\mathrm{H_2O_2}$

Molar mass of $\mathrm{H_2O_2} = 34$ g/mol, so

$m = \frac{1}{224}\times 34 = \frac{34}{224} \approx 0.152\ \text{g}$ …
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Q.A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL $^{-1}$ , then what shall be the molarity of the solution?

Concept understanding — Molality Calculation

Molality: The Concentration That Ignores Temperature

The Intuition

The Precise Definition

Worked Example

Common Mistake to Avoid

When Do You Use Molality?

Why this formula?

Molality Calculation: Why the Formula Works

The Definition First

Why Mass of Solvent, Not Solution?

The Reasoning

Deriving the Formula Step-by-Step

Step 1: Moles of Solute

Step 2: Mass of Solvent in kg

Step 3: Putting It Together

The Final Formula (Exam-Ready)

Why the ×1000 Factor? …

1. Molality — moles of solute per kg of solvent

2. Mole fraction of each component

Method: Mass-Based Composition Conversion (w/w % → Molality → Mole Fraction → Molarity)

Step 1: Interpret the 10% w/w label

Step 2: Calculate molality ( $m$ )

Step 3: Calculate mole fraction of each component

Step 4: Calculate molarity ( $M$ )

🧠 Common Mistake #1: Confusing % w/w with % w/v or % v/v

🧠 Common Mistake #2: Using density at the wrong step

🧠 Common Mistake #3: Forgetting to convert grams to kg for molality

🧠 Common Mistake #4: Wrong mole fraction formula

🧠 Common Mistake #5: Using density of pure water instead of solution density for molarity

Why this approach works

Step-by-step reasoning

Q.A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL−1^{-1}−1, then what shall be the molarity of the solution?

1. Molality — moles of solute per kg of solvent

2. Mole fraction of each component

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Q.A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL $^{-1}$ , then what shall be the molarity of the solution?

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