One mole of any substance contains $6.022 \times 10^{23}$ atoms/molecules. Number of molecules of $H_2SO_4$ present in 100 mL of 0.02M $H_2SO_4$ solution is ______. (i) $12.044 \times 10^{20}$ molecules (ii) $6.022 \times 10^{23}$ molecules (iii) $1 \times 10^{23}$ molecules (iv) $12.044 \times 10^{23}$ molecules

One mole of any substance contains 6.022 × 10²³ atoms/molecules. Number of molecules of H₂SO₄ present in 100 m

NCERT Exemplar · Q8

6.022 \times 10^{23}

(i)

12.044 \times 10^{20}

molecules

(ii)

6.022 \times 10^{23}

molecules

(iii)

1 \times 10^{23}

molecules

(iv)

12.044 \times 10^{23}

molecules

Telangana TsbieMCQ· 1mImportance★★★★★

58% · 52/89 Questions

🔒 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 — 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? …

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

Use $n = M \times V$ to find moles in the solution, then multiply by Avogadro's number to convert moles to molecules. The answer is $12.044 \times 10^{20}$ molecules.

The question asks for the number of molecules, not the amount in moles. Whenever you need to count individual particles—atoms, molecules, ions—you bridge from the macroscopic world (moles, concentration, volume) to the microscopic one using Avogadro's number, $N_A = 6.022 \times 10^{23} \, \text{mol}^{-1}$ .

The strategy is straightforward: first calculate how many moles of $H_2SO_4$ are present in the given volume of solution, then convert that amount to molecules.

Step-by-step calculation

Identify what you know.

The solution is 0.02 M (molarity), meaning there are 0.02 moles of $H_2SO_4$ per litre of solution. The volume given is 100 mL, which is $\frac{100}{1000} = 0.1$ L.
Calculate the number of moles.

Molarity is defined as moles per litre:

M = \frac{n}{V} \quad \Rightarrow \quad n = M \times V

Substituting the values:

n = 0.02 \, \text{mol/L} \times 0.1 \, \text{L} = 0.002 \, \text{mol}

So the solution contains $2 \times 10^{-3}$ moles of $H_2SO_4$ .

Convert moles to molecules. …

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

🧠 Concept Recap

We are given:

Volume = 100 mL = 0.1 L
Molarity = 0.02 M = 0.02 mol/L
Avogadro’s number = $6.022 \times 10^{23}$ molecules/mol

Correct approach:

Find moles of $H_2SO_4$ :

\text{Moles} = \text{Molarity} \times \text{Volume (in L)} = 0.02 \times 0.1 = 0.002 \text{ mol}

Find number of molecules:

\text{Molecules} = 0.002 \times 6.022 \times 10^{23} = 1.2044 \times 10^{21}

Which equals $12.044 \times 10^{20}$ molecules → Option (i).

✗ Common Mistake #1: Forgetting to convert mL to L

What students do:

They plug in 100 mL directly:

0.02 \times 100 = 2 \text{ moles}

Then multiply by Avogadro’s number → huge wrong answer like $12.044 \times 10^{23}$ (option (iv)).

Why it’s wrong:

Molarity is mol/L, so volume must be in litres.

✓ How to avoid:

Always write the unit:

\text{Volume in L} = \frac{\text{Volume in mL}}{1000}

Make it a habit — never skip this step.

✗ Common Mistake #2: Confusing atoms with molecules

What students do:

They calculate molecules correctly but then multiply by the number of atoms per molecule (e.g., 7 atoms in $H_2SO_4$ ) and give an answer for atoms, not molecules.

Why it’s wrong:

The question explicitly asks for number of molecules, not atoms.

✓ How to avoid:

Read the question carefully — underline “molecules” or “atoms”. If it says molecules, stop after multiplying moles by Avogadro’s number.

✗ Common Mistake #3: Misplacing the decimal in scientific notation

What students do:

They compute $0.002 \times 6.022 \times 10^{23}$ and write $1.2044 \times 10^{21}$ but then match it to $12.044 \times 10^{20}$ incorrectly, thinking it’s a different number.

Why it’s wrong:

$12.044 \times 10^{20}$ is exactly equal to $1.2044 \times 10^{21}$ — just a different notation.

✓ How to avoid:

Practice converting between forms: …

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

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 $_2$ . …
Start your 14-day free trial to see the full solution →
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.

$\text{Mass of pure HCl} = 4 \times 36.5 = 146 \text{ g}$ …
Start your 14-day free trial to see the full solution →
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) …
Start your 14-day free trial to see the full solution →
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

$m = \frac{1}{224}\times 34 = \frac{34}{224} \approx 0.152\ \text{g}$ …
Start your 14-day free trial to see the full solution →

←Previous questionQ7 — What will be the molality of the solution containing 18.25 g of HCl ga…→Next questionQ9 — What is the mass percent of carbon in carbon dioxide? (i) 0.034% (ii) …

Q.One mole of any substance contains $6.022 \times 10^{23}$ atoms/molecules. Number of molecules of $H_2SO_4$ present in 100 mL of 0.02M $H_2SO_4$ solution is ______.

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? …

Step-by-step calculation

Concept: Molarity and Avogadro's Number

Method: Mole–Molarity–Particle Conversion

Steps

🧠 Concept Recap

✗ Common Mistake #1: Forgetting to convert mL to L

✗ Common Mistake #2: Confusing atoms with molecules

✗ Common Mistake #3: Misplacing the decimal in scientific notation

Why this approach works

Step-by-step reasoning

Q.One mole of any substance contains 6.022×10236.022 \times 10^{23}6.022×1023 atoms/molecules. Number of molecules of H2SO4H_2SO_4H2​SO4​ present in 100 mL of 0.02M H2SO4H_2SO_4H2​SO4​ solution is ______.

Step-by-step calculation

🎓Unlock everything free for 14 days

Q.One mole of any substance contains $6.022 \times 10^{23}$ atoms/molecules. Number of molecules of $H_2SO_4$ present in 100 mL of 0.02M $H_2SO_4$ solution is ______.

Unlock everything free for 14 days