Which of the following statements indicates that law of multiple proportion is being followed. (i) Sample of c

NCERT Exemplar · Q15

Q. Which of the following statements indicates that law of multiple proportion is being followed.

(i) Sample of carbon dioxide taken from any source will always have carbon and oxygen in the ratio 1:2.

(ii) Carbon forms two oxides namely

CO_2

and CO, where masses of oxygen which combine with fixed mass of carbon are in the simple ratio 2:1.

(iii) When magnesium burns in oxygen, the amount of magnesium taken for the reaction is equal to the amount of magnesium in magnesium oxide formed.

(iv) At constant temperature and pressure 200 mL of hydrogen will combine with 100 mL oxygen to produce 200 mL of water vapour.

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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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The law of multiple proportion applies when two elements form more than one compound, and the masses of one element that combine with a fixed mass of the other are in a simple whole-number ratio. Option (ii) is the correct statement.

The law of multiple proportion is one of the fundamental laws of chemical combination, proposed by John Dalton. It deals specifically with pairs of elements that form more than one compound. The key idea is this: if element A and element B form two different compounds, the mass of B that combines with a fixed mass of A in the first compound, compared to the mass of B that combines with the same fixed mass of A in the second compound, will be in a ratio of small whole numbers (like 2:1, 3:2, etc.).

Why does this happen? Because atoms combine in fixed, whole-number ratios. If carbon and oxygen can form both CO and CO₂, then for the same number of carbon atoms, the oxygen atoms are in a 1:2 ratio — exactly what the law predicts.

Now let’s examine each option carefully.

Option (i) talks about a sample of carbon dioxide always having carbon and oxygen in the ratio 1:2. This is actually the law of constant proportion (or definite proportion) — that a given compound always contains the same elements in the same proportion by mass. It does not involve two different compounds, so it cannot illustrate the law of multiple proportion.
Option (ii) explicitly states that carbon forms two oxides, CO₂ and CO, and that the masses of oxygen combining with a fixed mass of carbon are in the simple ratio 2:1. This is a textbook example of the law of multiple proportion. Let’s verify: in CO, 12 g of carbon combines with 16 g of oxygen. In CO₂, 12 g of carbon combines with 32 g of oxygen. The ratio of oxygen masses is 16:32 = 1:2 (or 2:1 depending on order). That’s a simple whole-number ratio — exactly what the law demands. …

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Concept: Law of Multiple Proportions

The Law of Multiple Proportions states:

When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers.

Method: Fixed-Mass Comparison Method

Steps:

Identify the two elements involved in the compounds.
Fix the mass of one element (usually the first element) to a constant value across all compounds.
Find the masses of the other element that combine with this fixed mass.
Take the ratio of these masses.
Check if the ratio is a simple whole number ratio (like 2:1, 3:2, etc.).

Applying to the options:

Option	Analysis	Follows Law?
(i)	Talks about constant composition (same ratio from any source) — this is Law of Definite Proportions, not multiple proportions.	✗

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Here are the common mistakes students make with this question and how to avoid them.

Mistake 1: Confusing the Law of Multiple Proportions with the Law of Constant (Definite) Proportions

The Mistake:

Students often pick option (i) because it mentions a fixed ratio (1:2) of elements in a compound. They see "ratio" and assume it fits.

Why it is wrong:

Option (i) describes the Law of Constant Proportions (also called the Law of Definite Proportions). This law states that a given chemical compound always contains its component elements in a fixed ratio by mass, regardless of its source. The Law of Multiple Proportions, however, deals with two different compounds formed from the same two elements.

How to Avoid:

Use a keyword filter: When you see "fixed ratio" or "same ratio from any source," think Constant Proportions.
Look for two compounds: The Law of Multiple Proportions always requires comparing at least two distinct compounds (e.g., CO and CO₂, or H₂O and H₂O₂). Option (i) only talks about one compound (carbon dioxide).

Mistake 2: Misinterpreting the "Simple Ratio" Condition

The Mistake:

Students may understand that two compounds are needed, but they fail to check the fixed mass of one element condition. They might pick an option that shows a ratio of masses, but not for a fixed mass of the other element.

Why it is wrong:

The law states: "When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in a ratio of small whole numbers."

How to Avoid:

Always identify the "fixed mass" element. In option (ii), the statement explicitly says "masses of oxygen which combine with fixed mass of carbon." This is the correct setup.
Check the ratio: For CO (carbon monoxide), 12 g of carbon combines with 16 g of oxygen. For CO₂, 12 g of carbon combines with 32 g of oxygen. The masses of oxygen (16 and 32) are in the ratio $16:32 = 1:2$ , which is a simple whole number ratio. This is the perfect example.

Mistake 3: Confusing the Law of Multiple Proportions with the Law of Conservation of Mass

The Mistake:

Students pick option (iii) because it talks about the mass of magnesium before and after the reaction being equal.

Why it is wrong:

Option (iii) describes the Law of Conservation of Mass: mass is neither created nor destroyed in a chemical reaction. It has nothing to do with comparing ratios of elements in different compounds.

How to Avoid:

Recognize the "before and after" pattern: If a statement talks about the mass of reactants equaling the mass of products, it is always the Law of Conservation of Mass.
Remember the focus: Multiple proportions is about comparing the composition of different compounds formed by the same elements, not about mass balance in a single reaction.

Mistake 4: Confusing the Law of Multiple Proportions with the Law of Gaseous Volumes (Gay-Lussac's Law)

The Mistake:

Students pick option (iv) because it shows a simple volume ratio (2:1:2) of gases.

Why it is wrong: …

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

KEAM 2026Set eng-2026-04214 marksMCQ
$2.0\ ms^{-2} = X\ km\ h^{-2}$
›Reveal solutionSolution

Converting, $2.0\ ms^{-2}=2.592\times10^4\ km\,h^{-2}$ ; keeping two significant figures gives $2.6\times10^4$ .

Conversion factor: $1\ ms^{-2}=\dfrac{10^{-3}\ km}{(1/3600\ h)^2}=\dfrac{3600^2}{1000}\ km\,h^{-2}=\dfrac{12960000}{1000}=12960\ km\,h^{-2}$ .

So $2.0\ ms^{-2}=2.0\times12960=25920\ km\,h^{-2}=2.592\times10^4$ . …
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KEAM 2025Set eng-2025-04264 marksMCQ
$Q. One torr is (A) 1 mm of Hg (B) 1 cm of Hg (C) 76 mm of Hg (D) 100 mm of Hg (E) 76 cm of Hg$
›Reveal solutionSolution

By definition, $1\ \text{torr}=1\ \text{mm Hg}$ .

The torr is a unit of pressure defined as $\dfrac{1}{760}$ of a standard atmosphere, which equals the pressure exerted by a $1\ \text{mm}$ colu …
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KEAM 2025Set pha-2025-0424A4 marksMCQ
$10^{-28}m^2$
›Reveal solutionSolution

The barn is a unit of area for nuclear cross-sections: $1\,\text{barn} = 10^{-28}\,m^2$ (equal to $100\,fm^2$ ). …

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KEAM 2024Set pha-2024-06104 marksMCQ
$Q.In the measurement of length 6 $\mu$m is equal to $x$ pm. Then the value of $x$ is (A) $1.5 \times 10^{-5}$ (B) $1.2 \times 10^{6}$ (C) $3 \times 10^{-6}$ (D) $6 \times 10^{6}$ (E) $2 \times 10^{-12}$$
›Reveal solutionSolution

Convert micrometres to picometres.

$1\,\mu m = 10^{-6}$ m and $1$ pm $= 10^{-12}$ m. …
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Q.Which of the following statements indicates that law of multiple proportion is being followed.

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