Match the following: (i) 88 g of CO₂ (ii) 6.022 × 10²³ molecules of H₂O (iii) 5.6 litres of O₂ at STP (iv) 96

NCERT Exemplar · Q36

Q. Match the following:

(i) 88 g of

CO_2

(ii)

6.022 \times 10^{23}

molecules of

H_2O

(iii) 5.6 litres of

O_2

at STP

(iv) 96 g of

O_2

(v) 1 mol of any gas
with

(a) 0.25 mol

(b) 2 mol

(d)

6.022 \times 10^{23}

molecules

(e) 3 mol

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Concept understanding — Unit Conversion

Unit Conversion: Why 1 Metre and 100 Centimetres Are the Same Thing

Imagine you're measuring the length of your desk. You pull out a ruler marked in centimetres and find it's 120 cm long. Your friend, using a metre stick, says it's 1.2 m. You're both right — you've just used different units to describe the same physical length.

That's the core idea: unit conversion is the process of changing how you express a quantity without changing the quantity itself.

The Intuition: Same Quantity, Different Labels

Think of a pizza. Whether you call it "one pizza" or "8 slices," the amount of pizza hasn't changed. You've just used a different unit (pizza vs. slice) to describe it.

Similarly, 1 metre and 100 centimetres are the same length — just like 1 pizza and 8 slices are the same amount. The number changes (1 becomes 100, or 1 becomes 8), but the actual thing being measured stays identical.

Note

This is the most important idea to hold onto: conversion changes the number, not the quantity. If you ever feel like the quantity has changed, you've made a mistake.

The Precise Statement

Unit conversion is the multiplication of a quantity by a conversion factor — a fraction equal to 1 — that cancels the old unit and introduces the new one.

A conversion factor looks like this:

\frac{\text{new unit}}{\text{old unit}} = 1

For length: $\frac{1 \text{ m}}{100 \text{ cm}} = 1$ and $\frac{100 \text{ cm}}{1 \text{ m}} = 1$ .

Why are these fractions equal to 1? Because 1 metre is 100 centimetres. The numerator and denominator describe the same physical length, so their ratio is exactly 1.

How to Convert: The Only Rule You Need

Multiply by a conversion factor that cancels the unit you have and leaves the unit you want.

Let's convert 120 cm to metres:

Start with what you have: $120 \text{ cm}$
Choose the conversion factor that has "cm" in the denominator (to cancel it) and "m" in the numerator: $\frac{1 \text{ m}}{100 \text{ cm}}$
Multiply:

120 \text{ cm} \times \frac{1 \text{ m}}{100 \text{ cm}} = \frac{120}{100} \text{ m} = 1.2 \text{ m}

The "cm" units cancel just like numbers do: $\frac{\text{cm}}{\text{cm}} = 1$ .

Tip

Always write the units explicitly. If the units don't cancel correctly, you've used the wrong conversion factor. This catches 90% of conversion mistakes.

The Reverse: Metres to Centimetres

Now convert 1.2 m to cm. This time, you want "cm" to remain and "m" to cancel. Use $\frac{100 \text{ cm}}{1 \text{ m}}$ :

1.2 \text{ m} \times \frac{100 \text{ cm}}{1 \text{ m}} = 1.2 \times 100 \text{ cm} = 120 \text{ cm}

Notice: when going from a larger unit (m) to a smaller unit (cm), the number gets larger (1.2 → 120). When going from smaller to larger, the number gets smaller (120 → 1.2). This is a useful sanity check.

Common Conversion Factors You'll Use

Quantity	Relationship	Conversion Factors
Length	1 m = 100 cm	$\frac{1 \text{ m}}{100 \text{ cm}}$ , $\frac{100 \text{ cm}}{1 \text{ m}}$
Mass	1 kg = 1000 g	$\frac{1 \text{ kg}}{1000 \text{ g}}$ , $\frac{1000 \text{ g}}{1 \text{ kg}}$
Time	1 h = 60 min	$\frac{1 \text{ h}}{60 \text{ min}}$ , $\frac{60 \text{ min}}{1 \text{ h}}$
Speed	1 km/h = $\frac{1000}{3600}$ m/s	$\frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}}$

Why this formula?

Dimensional Analysis: Why the Key Principles Hold

Dimensional Analysis is a powerful tool in physics and engineering that lets us check the consistency of equations, derive relationships, and convert units. But why does it work? Let's build the reasoning from the ground up.

1. The Core Idea: Physical Quantities Have Dimensions

Every physical quantity (like length, time, mass) can be expressed in terms of fundamental dimensions. The most common set in mechanics is:

L = Length
M = Mass
T = Time

For example:

Speed has dimensions $[L T^{-1}]$
Force has dimensions $[M L T^{-2}]$
Energy has dimensions $[M L^2 T^{-2}]$

Why this matters: Two quantities can only be meaningfully compared or equated if they have the same dimensions. You cannot add apples to oranges — and you cannot add length to time.

2. The Principle of Dimensional Homogeneity

The key formula that underpins everything is:

Every valid physical equation must be dimensionally homogeneous.

This means: the dimensions on the left-hand side must equal the dimensions on the right-hand side.

Why must this hold?

Consider an equation like:

v = u + at

Left side: $[v] = L T^{-1}$
Right side: $[u] = L T^{-1}$ , $[at] = (L T^{-2})(T) = L T^{-1}$

Both sides have dimensions $L T^{-1}$ . If they didn't match, the equation would be physically meaningless — you'd be comparing quantities that cannot be equal in any real experiment.

Reasoning: Physical laws describe relationships between measurable quantities. If the dimensions don't match, the equation cannot represent a real physical relationship, because the numerical value would depend on the arbitrary choice of units.

3. The Buckingham Pi Theorem: Why We Can Derive Relationships

This is the deeper mathematical reason. The Buckingham Pi Theorem states:

If a physical problem involves $n$ variables and $k$ fundamental dimensions, then it can be reduced to $n-k$ independent dimensionless groups (called $\pi$ groups).

Why does this work?

Imagine you have a relationship:

f(Q_1, Q_2, \dots, Q_n) = 0

where each $Q_i$ has dimensions. Because the equation must be dimensionally homogeneous, we can rearrange it into a function of dimensionless products only:

F(\pi_1, \pi_2, \dots, \pi_{n-k}) = 0

The reasoning: Dimensions act as constraints. Each fundamental dimension (M, L, T) gives one constraint. So if you have $n$ variables and $k$ constraints, you only have $n-k$ independent dimensionless combinations.

Example: For a simple pendulum, the period $T$ depends on length $L$ , mass $m$ , and gravity $g$ . That's 4 variables with 3 dimensions (M, L, T). So $4-3 = 1$ dimensionless group: $\pi = \frac{T^2 g}{L}$ . This tells us $T \propto \sqrt{L/g}$ without solving any differential equation.

4. Why We Can Convert Units Using Dimensional Analysis

The conversion factor formula:

\text{Value in new unit} = \text{Value in old unit} \times \left(\frac{\text{old unit}}{\text{new unit}}\right)^{\text{dimension exponent}}

Why this works: …

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This matching problem is about converting mass, number of molecules, and volume at STP into moles, then pairing equal quantities. The correct matches are: (i)→(b), (ii)→(c), (iii)→(a), (iv)→(e), (v)→(d).

The entire exercise hinges on one idea: mole is the bridge between mass, number of particles, and gas volume. Once you convert every given quantity into moles, matching becomes trivial — you're just pairing equal numbers.

Let’s go through each item one by one.

(i) 88 g of CO₂

Molar mass of CO₂ = $12 + 2 \times 16 = 44$ g/mol.

Moles = $\frac{88}{44} = 2$ mol.

So (i) matches with (b) 2 mol.
(ii) $6.022 \times 10^{23}$ molecules of H₂O

That number is Avogadro’s constant — exactly 1 mole of anything.

So (ii) matches with (c) 1 mol.
(iii) 5.6 litres of O₂ at STP

At STP, 1 mole of any gas occupies 22.4 L.

Moles = $\frac{5.6}{22.4} = 0.25$ mol.

So (iii) matches with (a) 0.25 mol.
(iv) 96 g of O₂

Molar mass of O₂ = $2 \times 16 = 32$ g/mol.

Moles = $\frac{96}{32} = 3$ mol.

So (iv) matches with (e) 3 mol.
(v) 1 mol of any gas

By definition, 1 mole of any substance contains $6.022 \times 10^{23}$ particles.

So (v) matches with (d) $6.022 \times 10^{23}$ molecules. …

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Common Mistakes in Mole Concept Matching Problems

Students often lose marks in matching problems like this because they rush through conversions. Here are the most frequent errors and how to avoid each.

Mistake 1: Confusing Mass with Number of Particles

The error: Treating 88 g of $CO_2$ as 1 mol because "88 looks like a round number."

Why it happens: Students memorise molar masses incorrectly or assume all gases have the same molar mass.

How to avoid: Always calculate molar mass explicitly:

$CO_2$ : $12 + (2 \times 16) = 44 \text{ g/mol}$
So 88 g = $\frac{88}{44} = \textbf{2 mol}$

Key rule: Never guess molar masses — calculate them every time.

Mistake 2: Forgetting STP Conditions for Gases

The error: Treating 5.6 L of $O_2$ as 0.25 mol but forgetting to state "at STP" or using wrong molar volume.

Why it happens: Students memorise "22.4 L = 1 mol" but forget it applies only at STP (0°C, 1 atm).

How to avoid:

At STP: 1 mol of any gas = 22.4 L
So 5.6 L = $\frac{5.6}{22.4} = \textbf{0.25 mol}$

Remember: If STP is not mentioned, you cannot use 22.4 L/mol.

Mistake 3: Mixing Up $O_2$ and $O$ in Molar Mass

The error: Using 16 g/mol for $O_2$ instead of 32 g/mol.

Why it happens: Students confuse atomic mass (O = 16) with molecular mass ( $O_2$ = 32).

How to avoid:

For 96 g of $O_2$ : Molar mass = $2 \times 16 = 32 \text{ g/mol}$
Moles = $\frac{96}{32} = \textbf{3 mol}$

Check: Always ask — is it atomic or molecular oxygen?

Mistake 4: Misinterpreting Avogadro's Number

The error: Thinking $6.022 \times 10^{23}$ molecules always equals 1 mol, but forgetting it applies to any substance.

Why it happens: Students treat it as a special property of certain substances.

How to avoid:

$6.022 \times 10^{23}$ molecules = 1 mol of any substance
So (ii) matches with (c) 1 mol and also with (d) $6.022 \times 10^{23}$ molecules

Key insight: This is a definition, not a calculation.

Mistake 5: Forgetting That 1 Mol of Any Gas Has the Same Number of Molecules …

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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 17 on this concept.

GUJCET 2026Set x1 markMCQ
$Q. Out of the following physical quantities which quantity has the same unit as that of Planck's constant? (A) moment of force (B) power (C) angular momentum (D) moment of inertia$
›Reveal solutionSolution

$[h]=\text{J·s}$ , identical to the unit of angular momentum.

Planck's constant appears in $E=h\nu$ , so $[h]=\dfrac{[E]}{[\nu]}=\dfrac{\text{J}}{\text{s}^{-1}}=\text{J·s}=\text{kg·m}^2\text{s}^{-1}$ .

Angular momentum $L=mvr$ has units $\text{kg}\cdot(\text{m/s})\cdot\text{m}=\text{kg·m}^2\text{s}^{-1}=\text{J·s}$ . …
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GUJCET 2025Set 031 markMCQ
$[L^{2}]$
›Reveal solutionSolution
[!TLDR]

Current sensitivity $=\theta/I$ ; since $\theta$ is dimensionless, its dimension is $[A^{-1}]$ .

Concept

The current sensitivity of a moving-coil galvanometer is the deflection produced per unit current, $S_I=\dfrac{\theta}{I}=\dfrac{NBA}{k}$ . An angular deflection is dimensionless.

Solution …
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GUJCET 2024Set 131 markMCQ
$Q. Js is the unit of ________ physical quantity. (A) Angular momentum (B) Work function (C) Moment of Inertia (D) Rydberg constant$
›Reveal solutionSolution

Angular momentum $L=I\omega$ has SI unit $\text{kg m}^2\text{s}^{-1} = \text{J·s}$ . …

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GUJCET 2023Set 091 markMCQ
$kg^{-1}s^{-2}A$
›Reveal solutionSolution
[!TLDR]

Reducing $\text{m}^2\text{V}^{-1}\text{s}^{-1}$ to fundamental units gives $\text{kg}^{-1}\text{s}^2\text{A}$ .

Concept

Mobility $\mu=\dfrac{v_d}{E}$ has units of $\dfrac{\text{m/s}}{\text{V/m}}=\text{m}^2\text{V}^{-1}\text{s}^{-1}$ . Convert the volt to base units to express $\mu$ in fundamental units.

Solution
$\text{V}=\frac{\text{J}}{\text{C}}=\frac{\text{kg}\,\text{m}^2\text{s}^{-2}}{\text{A}\,\text{s}}=\text{kg}\,\text{m}^2\text{A}^{-1}\text{s}^{-3}.$
Then …
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GUJCET 2023Set 091 markMCQ
$M^1L^1T^{-2}A^{-2}$
›Reveal solutionSolution
[!TLDR] $[L]=M^1L^2T^{-2}A^{-2}$ → (B).

Concept

Self-inductance is defined by $\varepsilon=-L\dfrac{dI}{dt}$ , so $L=\dfrac{\varepsilon\,dt}{dI}$ . (NCERT Electromagnetic Induction.)

Solution …
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GUJCET 2023Set 091 markMCQ
$Q. The earth takes 24 h to rotate once about its axis. How much time does the Sun takes to shift by 1 minute viewed from the earth. (A) 4 minutes (B) 40 s (C) 4 s (D) 40 minutes$
›Reveal solutionSolution
[!TLDR]

The Sun moves 1° in 4 minutes, so 1 arcminute of apparent shift takes 4 seconds.

Concept

Earth's rotation makes the Sun appear to sweep $360^\circ$ in 24 hours; a small angular shift corresponds to a proportional time.

Solution

Angular rate of the Sun's apparent motion:

$\frac{360^\circ}{24\ \text{h}} = 15^\circ/\text{h} = \frac{1^\circ}{4\ \text{min}}$ …
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GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. The dimensional formula of electric flux is ___.$
(a) M^1 L^-3 T^-3 A^-1
(b) M^1 L^3 T^-3 A^-1
(c) M^-1 L^3 T^-3 A^-1
(d) M^1 L^3 T^3 A^-1
›Reveal solutionSolution

Flux = E x area, i.e. (V/m) x m^2 = volt-metre; expressing volt in base units gives M L^3 T^-3 A^-1.

Electric flux Phi = E x A, with units (V/m)(m^2) = V.m.

…

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GUJCET 2021Set 151 markMCQ
$[M^{-1} L^{2} T^{-3} A^{1}]$
›Reveal solutionSolution

Electric potential is energy per unit charge.

Concept: …
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GUJCET 2020Set 071 markMCQ
$1^\circ$
›Reveal solutionSolution

$24$ hours per $360^\circ$ gives $4$ minutes per degree.

Concept: Earth turns $360^\circ$ in $24$ h, so the angular rate is $360^\circ/24\,\text{h}=15^\circ/\text{h}$ . …
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GUJCET 2020Set 071 markMCQ
$V\cdot s\cdot A^{-1}$
›Reveal solutionSolution

$1\,\text{H}=\text{Wb/A}=\text{V·s/A}$ ; $\text{Wb·s·A}^{-1}$ has an extra second and is not inductance. …

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GUJCET 2019Set 131 markMCQ
$M^0L^0T^0$
›Reveal solutionSolution

Torsional constant has the dimensions of torque, $M^1L^2T^{-2}$ .

Concept: For a torsion spring, restoring torque $\tau = C\theta$ , so $C = \tau/\theta$ . Angle $\theta$ is dimensionless, hence $C$ has the dimensions of torque (energy).

Steps: …
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GUJCET 2019Set 131 markMCQ
$M^1L^2T^1Q^{-2}$
›Reveal solutionSolution
The dimensional formula asked for is that of resistance (equivalently inductive reactance $\omega L$ ): $M^1L^2T^{-1}Q^{-2}$ .

Concept: Resistance $R=\dfrac{V}{I}$ . Expressing potential and current in terms of charge $Q$ : $V=\dfrac{\text{energy}}{Q}=\dfrac{ML^2T^{-2}}{Q}$ and $I=\dfrac{Q}{T}$ .

Steps:

$[R]=\dfrac{[V]}{[I]}=\dfrac{ML^2T^{-2}Q^{-1}}{Q\,T^{-1}}$ . …
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←Previous questionQ35 — The reactant which is entirely consumed in reaction is known as limiti…→Next questionQ37 — Match the following physical quantities with units: Physical quantity:…

Q.Match the following:

Concept understanding — Unit Conversion

Unit Conversion: Why 1 Metre and 100 Centimetres Are the Same Thing

The Intuition: Same Quantity, Different Labels

The Precise Statement

How to Convert: The Only Rule You Need

The Reverse: Metres to Centimetres

Common Conversion Factors You'll Use

Why this formula?

Dimensional Analysis: Why the Key Principles Hold

1. The Core Idea: Physical Quantities Have Dimensions

2. The Principle of Dimensional Homogeneity

Why must this hold?

3. The Buckingham Pi Theorem: Why We Can Derive Relationships

Why does this work?

4. Why We Can Convert Units Using Dimensional Analysis

Why this works: …

Concept: Mole Concept and Stoichiometric Conversions

Method: Unit Conversion to Moles

(i) 88 g of $CO_2$

(ii) $6.022 \times 10^{23}$ molecules of $H_2O$

(iii) 5.6 litres of $O_2$ at STP

Common Mistakes in Mole Concept Matching Problems

Mistake 1: Confusing Mass with Number of Particles

Mistake 2: Forgetting STP Conditions for Gases

Mistake 3: Mixing Up $O_2$ and $O$ in Molar Mass

Mistake 4: Misinterpreting Avogadro's Number

Mistake 5: Forgetting That 1 Mol of Any Gas Has the Same Number of Molecules …

Concept

Solution …

Concept

Solution

Concept

Solution …

Concept

Solution

Q.Match the following:

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