Assertion (A): One atomic mass unit is defined as one twelfth of the mass of one carbon-12 atom. Reason (R): C

NCERT Exemplar · Q39

Q. Assertion (A): One atomic mass unit is defined as one twelfth of the mass of one carbon-12 atom. Reason (R): Carbon-12 isotope is the most abundunt isotope of carbon and has been chosen as standard. Choose the correct option out of the choices given below.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

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Concept understanding — Molecular Mass Calculation

What is Molecular Mass? The Intuition

Imagine you're at a market buying apples. You don't weigh each apple individually — you put a dozen on a scale. The total weight tells you something about the apples, but it also depends on how many apples you have.

Atoms and molecules are unimaginably tiny. A single water molecule ( $H_2O$ ) weighs about $3 \times 10^{-23}$ grams. That number is useless for practical chemistry. So instead of working with individual molecules, chemists count them in huge fixed numbers — specifically, $6.022 \times 10^{23}$ molecules, which is called one mole.

Molecular mass is simply the mass of one mole of a substance, expressed in grams per mole (g/mol). It answers the question: "If I have $6.022 \times 10^{23}$ molecules of this compound, how much would they weigh on a lab balance?"

Note

The number $6.022 \times 10^{23}$ is Avogadro's constant. It's chosen so that the mass of one mole of carbon-12 atoms is exactly 12 grams — matching the atomic mass unit scale perfectly.

The Precise Definition

Molecular mass (also called molar mass) is the mass of one mole of a molecular substance. It is numerically equal to the sum of the atomic masses of all atoms in the molecule, expressed in g/mol.

For example:

Water ( $H_2O$ ): 2 hydrogen atoms + 1 oxygen atom
Atomic mass of H = 1.008 g/mol
Atomic mass of O = 16.00 g/mol
Molecular mass of $H_2O$ = $2(1.008) + 16.00 = 18.016$ g/mol

\text{Molecular mass} = \sum (\text{number of atoms of each element} \times \text{atomic mass of that element})

How to Calculate It — Step by Step

Let's take glucose, $C_6H_{12}O_6$ , as a worked example.

Step 1: Identify each element and its count

Carbon (C): 6 atoms
Hydrogen (H): 12 atoms
Oxygen (O): 6 atoms

Step 2: Look up atomic masses (from the periodic table)

C: 12.01 g/mol
H: 1.008 g/mol
O: 16.00 g/mol

Step 3: Multiply and add

\text{Molecular mass} = 6(12.01) + 12(1.008) + 6(16.00)

= 72.06 + 12.096 + 96.00

= 180.156 \text{ g/mol}

Tip

Always keep at least 2 decimal places from the periodic table. For exam problems, they usually give you atomic masses — use exactly what's provided.

Why This Matters

Molecular mass is the bridge between the microscopic world (atoms and molecules) and the macroscopic world (grams you can weigh). Once you know the molecular mass, you can:

Convert grams to moles: $\text{moles} = \frac{\text{mass in grams}}{\text{molecular mass}}$
Convert moles to grams: $\text{mass} = \text{moles} \times \text{molecular mass}$
Determine the number of molecules: $\text{molecules} = \text{moles} \times 6.022 \times 10^{23}$

Watch out

Do not confuse molecular mass with atomic mass. Atomic mass refers to a single element (like oxygen = 16.00 g/mol). Molecular mass refers to a compound (like $CO_2$ = 44.01 g/mol). Also, for ionic compounds like NaCl, we use formula mass (same calculation, but the substance isn't molecular).

Common Exam Pitfalls

Forgetting to multiply by the subscript. In $H_2SO_4$ , there are 2 hydrogens, not 1. …

Why this formula?

Stoichiometry & Mole Calculation: The "Why" Behind the Formula

Let's build this from the ground up — not as a list of formulas to memorise, but as a logical chain of reasoning.

1. The Core Question: What is a Mole?

A mole is simply a counting unit, like a dozen (12) or a gross (144). But instead of 12, a mole contains $6.022 \times 10^{23}$ particles (Avogadro's number, $N_A$ ).

Why this number?

It was chosen so that 1 mole of any substance has a mass in grams equal to its atomic/molecular mass in amu.

Example: 1 atom of carbon-12 has mass 12 amu.
1 mole of carbon-12 has mass 12 grams.

This is the bridge between the microscopic (atoms/molecules) and the macroscopic (grams we can weigh).

2. The Fundamental Relationship

The key formula is:

n = \frac{m}{M}

Where:

$n$ = number of moles
$m$ = mass of substance (in grams)
$M$ = molar mass (in g/mol)

Why does this work?

Think of it as a conversion factor:

If 1 mole of a substance weighs $M$ grams, then $m$ grams contains $\frac{m}{M}$ moles.

Derivation logic:

Molar mass $M$ tells you: "1 mol = $M$ g"
So the conversion factor is $\frac{1 \text{ mol}}{M \text{ g}}$
Multiply mass $m$ by this factor: $m \times \frac{1}{M} = \frac{m}{M}$ moles

3. Connecting to Number of Particles

n = \frac{N}{N_A}

Where:

$N$ = number of particles (atoms, molecules, ions)
$N_A = 6.022 \times 10^{23}$ particles/mol

Why?

1 mole = $N_A$ particles
So $N$ particles = $\frac{N}{N_A}$ moles

Combine both formulas:

\frac{m}{M} = \frac{N}{N_A}

This single equation ties mass, molar mass, number of particles, and Avogadro's number together.

4. The Gas Volume Connection (for gases at STP)

For gases only:

n = \frac{V}{22.4 \text{ L/mol}}

Why 22.4 L?

From the ideal gas law: $PV = nRT$

At STP (Standard Temperature and Pressure: 0°C, 1 atm):

$P = 1$ atm
$T = 273.15$ K
$R = 0.0821$ L·atm/(mol·K)

For $n = 1$ mole:

V = \frac{nRT}{P} = \frac{1 \times 0.0821 \times 273.15}{1} \approx 22.4 \text{ L}

So 1 mole of any ideal gas occupies 22.4 L at STP. This is a consequence of the gas laws, not a definition.

5. The Stoichiometry Chain: From One Substance to Another

In a balanced chemical equation like:

aA + bB \rightarrow cC + dD

The coefficients tell you the mole ratio:

\frac{\text{moles of A}}{\text{moles of B}} = \frac{a}{b}

Why this works: …

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The assertion is correct because the definition of 1 amu is exactly $\frac{1}{12}$ the mass of a carbon-12 atom. The reason is also correct — carbon-12 is the standard — but it does not explain why the definition uses one-twelfth; it only states a fact about abundance. Hence both are true, but (R) is not the correct explanation of (A).

The question tests your understanding of the definition of the atomic mass unit and the logic behind choosing carbon-12 as the standard. Many students memorise the definition but miss the nuance of what counts as a "correct explanation" in assertion-reason problems.

Let’s break it down.

What is an atomic mass unit (amu)?

It is a unit of mass used to express atomic and molecular masses. By international agreement (since 1961), 1 amu is defined as exactly one-twelfth of the mass of a single carbon-12 atom. This is a fixed, absolute definition — it does not depend on abundance or any other isotope.

Why carbon-12?

Before 1961, chemists used oxygen-16 as the standard, and physicists used a different one, causing confusion. Carbon-12 was chosen because:

It is stable and easy to work with.
It gives nearly integer mass numbers for most elements (e.g., hydrogen ≈ 1, oxygen ≈ 16).
It is not the most abundant isotope of carbon (carbon-12 is about 98.9% abundant, but carbon-13 also exists). The reason says "most abundant" — that is actually true, but it is not the reason the definition uses one-twelfth.

The real reason for the one-twelfth fraction is convenience: it makes the mass of a carbon-12 atom exactly 12 amu, so that the mass of a proton or neutron is roughly 1 amu. The abundance of carbon-12 is a separate fact, not the logical basis for the definition.

Now, examine the two statements:

Assertion (A): One atomic mass unit is defined as one twelfth of the mass of one carbon-12 atom. Reason (R): Carbon-12 isotope is the most abundant isotope of carbon and has been chosen as standard.

Is (A) true? Yes. This is the exact definition from the IUPAC and SI systems. No ambiguity.
Is (R) true? Yes. Carbon-12 is indeed the most abundant carbon isotope (≈98.9%). It was chosen as the standard in 1961. …

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Here’s a breakdown of the common mistakes students make on this specific Assertion-Reason question, along with how to avoid each.

The Correct Answer First

Assertion (A): True.
Reason (R): True.
Relationship: R is not the correct explanation of A.

Correct Option: (B) — Both A and R are true, but R is not the correct explanation of A.

Common Mistake #1: Thinking the Reason is False

The Mistake: Students often think the Reason is false because they believe carbon-12 is not the most abundant isotope. They recall that carbon-12 is the standard, but they confuse abundance with the reason for its selection.

Why it’s wrong: Carbon-12 is the most abundant isotope of carbon (about 98.9% of natural carbon). The statement in R is factually correct.

How to Avoid:

Memorise the facts separately:
- Carbon-12 is the most abundant carbon isotope.
- Carbon-12 was chosen as the standard not because of abundance, but because it is stable, easy to measure, and gives whole-number atomic masses for most elements.
Don’t assume: Just because something is a standard doesn’t mean it’s chosen for the reason you think. Always check the exact wording of the Reason.

Common Mistake #2: Choosing “Both A and R are true, and R is the correct explanation of A”

The Mistake: Students see that both statements are true and automatically assume the Reason explains the Assertion. They think: “Carbon-12 is abundant, so it was chosen as the standard — makes sense.”

Why it’s wrong: The Assertion is about the definition of the atomic mass unit (1 amu = 1/12th mass of one C-12 atom). The Reason talks about why C-12 was chosen as the standard.

The actual reason for choosing C-12 is not its abundance. It was chosen because it is a stable, non-radioactive isotope that gives nearly whole-number atomic masses for most elements, making calculations simpler.
Abundance is a coincidental fact, not the causal explanation.

How to Avoid:

Ask yourself: “Does the Reason directly cause the Assertion to be true?”
- Here: Does the abundance of C-12 cause the definition of amu to be 1/12th of its mass? No. The definition is a convention set by scientists, not a consequence of abundance.
Learn the real reason: The standard was shifted from oxygen-16 to carbon-12 in 1961 to unify physics and chemistry scales. Abundance was not the deciding factor.

Common Mistake #3: Misreading the Assertion as False

The Mistake: Some students think the definition of amu is “one twelfth of the mass of a carbon-12 atom” — which is correct — but they confuse it with the mass number or molar mass. …

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

Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2023Set ANNUAL2 marks
$Q. Pressure of 1g of an ideal gas 'A' at 27°C is found to be 2 bar. When 2g of another ideal gas 'B' is introduced in the same flask at the same temperature, the pressure becomes 3 bar. Find a relationship between their molecular masses.$
›Reveal solutionSolution

Setting up PV = nRT for each gas separately and dividing the equations gives M_B = 4 x M_A.

Given: 1 g of ideal gas A at 27 deg C (300 K) gives a pressure of 2 bar in a flask of volume V. Then 2 g of ideal gas B is added at the SAME temperature and volume, and the total pressure becomes 3 bar.

Step 1 -- Partial pressure of B (Dalton's Law of partial pressures):

P(total) = P(A) + P(B)

3 = 2 + P(B) => P(B) = 1 bar

Step 2 -- Apply the ideal gas equation to gas A:

P(A) V = n(A) R T = (1 g / M_A) R T

2V = (1/M_A) R T ... (i)

Step 3 -- Apply the ideal gas equation to gas B: …

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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2020Set ANNUAL2 marks
$Q. At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?$
›Reveal solutionSolution

Using density = PM/RT at the same temperature for both gases, equal densities mean P1·M1 = P2·M2; solving with the known values gives a molar mass of 70 g/mol for the oxide.

From the ideal gas equation, PV = nRT = (mass/M)RT, so density (d = mass/V) is:

d = PM/RT

Given both gases are at the same temperature (0°C = 273 K):

For the oxide: d_oxide = P_oxide × M_oxide / RT, with P_oxide = 2 bar

For N2: d_N2 = P_N2 × M_N2 / RT, with P_N2 = 5 bar, M_N2 = 28 g/mol

…

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Q.Assertion (A): One atomic mass unit is defined as one twelfth of the mass of one carbon-12 atom.
Reason (R): Carbon-12 isotope is the most abundunt isotope of carbon and has been chosen as standard.
Choose the correct option out of the choices given below.

Concept understanding — Molecular Mass Calculation

What is Molecular Mass? The Intuition

The Precise Definition

How to Calculate It — Step by Step

Why This Matters

Common Exam Pitfalls

Why this formula?

Stoichiometry & Mole Calculation: The "Why" Behind the Formula

1. The Core Question: What is a Mole?

2. The Fundamental Relationship

Why does this work?

3. Connecting to Number of Particles

Why?

4. The Gas Volume Connection (for gases at STP)

Why 22.4 L?

5. The Stoichiometry Chain: From One Substance to Another

Why this works: …

Concept: Atomic Mass Unit and Standard Isotope Selection

Method: Verification of Assertion-Reason Relationship

The Correct Answer First

Common Mistake #1: Thinking the Reason is False

Common Mistake #2: Choosing “Both A and R are true, and R is the correct explanation of A”

Common Mistake #3: Misreading the Assertion as False

Q.Assertion (A): One atomic mass unit is defined as one twelfth of the mass of one carbon-12 atom. Reason (R): Carbon-12 isotope is the most abundunt isotope of carbon and has been chosen as standard. Choose the correct option out of the choices given below.

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Q.Assertion (A): One atomic mass unit is defined as one twelfth of the mass of one carbon-12 atom.
Reason (R): Carbon-12 isotope is the most abundunt isotope of carbon and has been chosen as standard.
Choose the correct option out of the choices given below.

Unlock everything free for 14 days