Write the volume of 4.4 gram carbon dioxide at STP.

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Q. Write the volume of 4.4 gram carbon dioxide at STP.

Rajasthan RbseRajasthan Board Senior Secondary Part-I Examination 2017Subjective· 1mImportance★★★★★

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Concept understanding — Mole Concept

The Intuition: Why Do We Need a "Mole"?

Imagine you run a bakery and need to buy eggs. You don't go to the shop and say "I want 12 eggs" — you say "I want a dozen eggs." The word dozen is just a convenient name for the number 12. It saves you from counting out every single egg.

Now think about chemistry. Atoms and molecules are unimaginably tiny. A single grain of sand contains about $10^{19}$ atoms. If you tried to count them one by one, you'd be counting for billions of years. So chemists needed a "dozen" — but for particles that are astronomically small. That's the mole.

Note

Just as 1 dozen = 12 things, 1 mole = $6.022 \times 10^{23}$ things. That number is called Avogadro's number ( $N_A$ ).

Why that specific number? Because it was chosen so that one mole of any substance has a mass in grams equal to its atomic/molecular mass in atomic mass units (u). For example:

One atom of carbon-12 has mass 12 u. One mole of carbon-12 atoms has mass exactly 12 grams.
One molecule of water ( $H_2O$ ) has mass 18 u. One mole of water molecules has mass exactly 18 grams.

This is the bridge between the invisible atomic world and the measurable laboratory world.

The Precise Definition

Important

One mole is the amount of a substance that contains exactly $6.02214076 \times 10^{23}$ elementary entities (atoms, molecules, ions, electrons, etc.). This number is Avogadro's constant, $N_A$ .

The mole is a counting unit, like a dozen or a gross. It tells you how many particles you have, not how heavy they are.

The Three Pillars of the Mole Concept

The mole connects three measurable quantities: mass, number of particles, and volume of a gas. Here's how.

1. Mass ↔ Moles ↔ Number of Particles

The molar mass ( $M$ ) of a substance is the mass of one mole of it, in grams per mole (g/mol).

\text{Number of moles } (n) = \frac{\text{Mass of substance (g)}}{\text{Molar mass (g/mol)}}

\text{Number of particles} = n \times N_A = \frac{\text{Mass}}{M} \times 6.022 \times 10^{23}

Example: How many atoms are in 24 g of carbon?

Molar mass of carbon = 12 g/mol.

$n = \frac{24}{12} = 2$ moles.

Number of atoms = $2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{24}$ atoms.

Tip

Always check: if you have a mass in grams, divide by the molar mass to get moles. Then multiply by $N_A$ to get particles.

2. Volume of a Gas ↔ Moles

For gases, there's a special shortcut. At Standard Temperature and Pressure (STP) — 0°C and 1 atm pressure — one mole of any gas occupies 22.4 litres. This is called the molar volume.

\text{Number of moles } (n) = \frac{\text{Volume of gas at STP (L)}}{22.4 \text{ L/mol}}

Watch out

This 22.4 L/mol applies only at STP. If temperature or pressure changes, the volume changes. Use the ideal gas law ( $PV = nRT$ ) for non-STP conditions.

Example: What is the volume of 2 moles of oxygen gas at STP?

Volume = $2 \times 22.4 = 44.8$ litres.

Putting It All Together: The Mole Triangle

You can visualise the relationships as a triangle:

\text{Mass (g)} \xrightarrow{\div M} \text{Moles} \xrightarrow{\times N_A} \text{Particles}

\text{Volume of gas at STP (L)} \xrightarrow{\div 22.4} \text{Moles}

Any problem in mole concept is just a matter of converting along these paths. You never need to memorise a hundred formulas — just these three conversions.

A Worked Example …

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4.4 g of CO2 = 0.1 mole, which occupies 2.24 L at STP.

Step 1: Find the molar mass of CO2.

M(CO2) = 12 + 2(16) = 44 g/mol.

Step 2: Find the number of moles in 4.4 g.

n = given mass / molar mass = 4.4/44 = 0.1 mol.

…

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

Showing the 12 most recent of 43 on this concept.

CBSE 2026Set sz1 markMCQ
$Q. Select the correct one: The number of molecules in 89.6 litre of a gas at NTP are:$
(a) 6.023 x 10^23
(b) 2 x 6.023 x 10^23
(c) 3 x 6.023 x 10^23
(d) 4 x 6.023 x 10^23
›Reveal solutionSolution

Moles = Volume at NTP / 22.4 L mol^-1; here that gives 4 moles, so molecules = 4 x 6.023 x 10^23.

At NTP (Normal Temperature and Pressure, taken as 0 degC and 1 atm), the molar volume of any ideal gas is 22.4 L/mol (Avogadro's law consequence).

Number of moles, n = Given volume / Molar volume at NTP

n = 89.6 L / 22.4 L mol^-1 = 4 mol

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. "At the same temperature and pressure, equal volumes of all gases contain equal number of molecules." This statement is based on which law ?$
(a) Berzelius' law
(b) Avogadro's hypothesis
(c) Graham's law
(d) Charles' law
›Reveal solutionSolution

This is Avogadro's hypothesis (law).

Avogadro (1811) proposed that equal volumes of all gases, at the same temperature and pressure, contain equal numbers of molecules. This explained Gay-Lussac's law of combining v …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. A group of 6.022 \times 10^23 particles is called$
(a) Mole
(b) Atomic mass unit
(c) Graham's law
(d) Charles' law
›Reveal solutionSolution

6.022 × 10^23 particles = 1 mole.

The mole is the SI unit for amount of substance. One mole of any species contains exactly Avogadro's number of particles, 6.022 × 10^23. So 6.0 …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The total number of moles in 720 gm of water is$
(a) 4
(b) 10
(c) 40
(d) 72
›Reveal solutionSolution

720 g of water = 720/18 = 40 moles.

Molar mass of water (H2O) = 2(1) + 16 = 18 g/mol. …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The necessary volume of oxygen that should be required to convert 10 ml of SO2 to SO3 by complete oxidation is$
(a) 10 ml
(b) 20 ml
(c) 30 ml
(d) 5 ml
›Reveal solutionSolution

2SO2 + O2 → 2SO3, so 10 mL SO2 needs 5 mL O2.

Balanced reaction: 2SO2(g) + O2(g) → 2SO3(g). By Gay-Lussac's law of combining volumes, the volume ratio equals the mole ratio: 2 volumes …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The volume of 4.4 gm of CO2 at standard temperature and pressure is$
(a) 22.4 L
(b) 11.2 L
(c) 5.6 L
(d) 2.24 L
›Reveal solutionSolution

4.4 g CO2 = 0.1 mol → 2.24 L at STP.

Molar mass of CO2 = 12 + 2(16) = 44 g/mol.

Moles = 4.4 / 44 = 0.1 mol. …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. What is the SI unit of the amount of chemical substance?$
(a) kilogram
(b) gram
(c) mole
(d) tonne
›Reveal solutionSolution

The SI unit for amount of substance is the mole (symbol mol).

The mole is defined as the amount of substance that contains as many elementary entities (atoms, molecules, ions, etc.) as there are atoms in exactly 12 g of carbon-12, i.e. Avogadro's number (6.022 …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following has the largest number of atoms?$
(a) 0.5 g atom of Cu
(b) 0.635 g of Cu
(c) 0.25 mole of Cu atom
(d) 1 g of Cu
›Reveal solutionSolution
0.5 g atom of Cu contains the most atoms (0.5 mol x N_A).

Convert each option into moles of Cu atoms:

(a) 0.5 g atom of Cu = 0.5 mol Cu atoms = 0.5 x 6.022x10^23 = 3.011x10^23 atoms.

(b) 0.635 g Cu / 63.5 g mol^-1 = 0.01 mol = 6.02x10^21 atoms.

(c) 0.25 mol Cu atoms = 0.25 x 6.022x10^23 = 1.5055x10^23 atoms. …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. How much of NaOH is required to neutralise 1500 cc of 0.1 N HCl?$
(a) 40 g
(b) 4 g
(c) 6 g
(d) 60 g
›Reveal solutionSolution

6 g of NaOH exactly neutralises 1500 mL of 0.1 N HCl.

Equivalents of HCl = Normality x Volume(L) = 0.1 x 1.5 = 0.15 equivalents.

At neutralisation, equivalents of NaOH required = 0.15.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The amount of CO2 that could be produced when one mole of carbon is burnt in air, is$
(a) 22 g
(b) 50 g
(c) 44 g
(d) 56 g
›Reveal solutionSolution

C + O2 -> CO2 is a 1:1 mole reaction, so 1 mole of carbon gives exactly 1 mole (44 g) of CO2.

Write the balanced combustion equation for carbon burning completely in air (excess oxygen):

C(s) + O2(g) -> CO2(g)

The stoichiometric coefficients show a 1:1:1 mole ratio between C, O2, and CO2. So 1 mole of carbon reacts with 1 mole of O2 to give exactly 1 mole of CO2.

…

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CBSE 2025Set sz1 markMCQ
$Q. Select the correct one: Which of the following contours (contains) maximum number of atoms?$
(a) 6.023 x 10^21 molecules of CO2
(b) 22.4 L of CO2 at STP
(c) 0.44 g of CO2
(d) None of these
›Reveal solutionSolution

22.4 L of CO2 at STP = 1 mole of CO2 = 3 moles of atoms, the largest amount among the four options.

Each CO2 molecule has 3 atoms (1 C + 2 O), so moles of atoms = 3 x moles of CO2.

(A) 6.023 x 10^21 molecules of CO2 = 6.023x10^21 / 6.022x10^23 = 0.01 mol CO2 -> 0.03 mol atoms.

(B) 22.4 L of CO2 at STP = 1 mol of gas (since 1 mole of any ideal gas occupies 22.4 L at STP) -> 1 mol CO2 -> 3 mol atoms.

(C) 0.44 g of CO2: molar mass of CO2 = 44 g/mol, so 0.44 g = 0.01 mol CO2 -> 0.03 mol atoms.

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The number of molecules in 4.25 gm ammonia is$
(a) 1.0 × 10^23
(b) 1.5 × 10^23
(c) 2.0 × 10^23
(d) 2.5 × 10^23
›Reveal solutionSolution

4.25 g of NH3 contains 0.25 mol, i.e. about 1.5 × 10²³ molecules.

Molar mass of NH3 = 14 (N) + 3×1 (H) = 17 g/mol.

Moles of NH3 = $\dfrac{4.25}{17} = 0.25$ mol.

…

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Q.Write the volume of 4.4 gram carbon dioxide at STP.

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