Match the following physical quantities with units: Physical quantity: (i) Molarity (ii) Mole fraction (iii) M

NCERT Exemplar · Q37

Q. Match the following physical quantities with units: Physical quantity:

(i) Molarity

(ii) Mole fraction

(iii) Mole

(iv) Molality

(v) Pressure

(vi) Luminous intensity

(vii) Density

(viii) Mass
Unit:

(a)

g\ mL^{-1}

(b) mol

(d) Unitless

(e)

mol\ L^{-1}

(f) Candela

(g)

mol\ kg^{-1}

(h)

Nm^{-1}

(i) kg

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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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Understanding the definition of each physical quantity is key to identifying its correct unit. Molarity is $mol\ L^{-1}$ , Mole fraction is unitless, Mole is $mol$ , Molality is $mol\ kg^{-1}$ , Pressure is Pascal, Luminous intensity is Candela, Density is $g\ mL^{-1}$ , and Mass is $kg$ .

When matching physical quantities to their units, the most reliable approach is to recall or derive the definition of each quantity. Units are not arbitrary; they directly reflect the mathematical relationship between the fundamental quantities that define a particular physical property. By understanding "what" a quantity measures, we can deduce "how" it is measured, which leads us to its unit.

Let's go through each physical quantity and determine its unit based on its definition.

Molarity (i)
- Definition: Molarity ( $M$ ) is defined as the number of moles of solute dissolved per liter of solution.
- Formula: $M = \frac{\text{moles of solute}}{\text{volume of solution (L)}}$
- Unit Derivation: From the formula, the unit will be moles divided by liters.
- Unit: $mol\ L^{-1}$
- Match: This corresponds to unit (e).
Mole fraction (ii)
- Definition: Mole fraction ( $X$ ) of a component in a mixture is the ratio of the number of moles of that component to the total number of moles of all components in the mixture.
- Formula: $X_A = \frac{\text{moles of component A}}{\text{total moles of all components}}$
- Unit Derivation: Since it is a ratio of moles to moles, the units cancel out.
- Unit: Unitless
- Match: This corresponds to unit (d).
Mole (iii)
- Definition: The mole ( $mol$ ) is the SI unit for the amount of substance. It represents a specific number of particles (Avogadro's number, $6.022 \times 10^{23}$ ).
- Unit: $mol$
- Match: This corresponds to unit (b).
Molality (iv)
- Definition: Molality ( $m$ ) is defined as the number of moles of solute dissolved per kilogram of solvent.
- Formula: $m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}}$
- Unit Derivation: From the formula, the unit will be moles divided by kilograms.
- Unit: $mol\ kg^{-1}$
- Match: This corresponds to unit (g).
Pressure (v)
- Definition: Pressure ( $P$ ) is defined as the force exerted perpendicularly on a surface per unit area.
- Formula: $P = \frac{\text{Force}}{\text{Area}}$
- Unit Derivation: The SI unit of force is Newton ( $N$ ), and the SI unit of area is square meter ( $m^2$ ). Therefore, the SI unit of pressure is $N/m^2$ , which is also known as Pascal ( $Pa$ ).
- Unit: Pascal ( $Pa$ ) or $Nm^{-2}$
- Match: This corresponds to unit (c).
  Watch out
  
  Unit (h) $Nm^{-1}$ is provided as an option. This is NOT a unit of pressure. Pressure is force per area ( $Nm^{-2}$ ), not force per length ( $Nm^{-1}$ ). $Nm^{-1}$ is a unit for quantities like surface tension. Be careful with similar-looking units.
Luminous intensity (vi) …

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Method: Dimensional Analysis & Definition Matching

This method works by recalling the definition of each physical quantity and then identifying the unit that matches its formula.

Steps:

Recall the definition of each physical quantity.
Write the formula in terms of base quantities (mass, length, time, amount of substance, luminous intensity).
Match the derived unit from the list.

Step-by-step matching:

Physical Quantity	Definition / Formula	Derived Unit	Match
(i) Molarity	Moles of solute per litre of solution	$\frac{\text{mol}}{\text{L}} = \text{mol L}^{-1}$	(e)
(ii) Mole fraction	Moles of one component ÷ total moles	Ratio → unitless	(d)
(iii) Mole	SI base unit for amount of substance	mol	(b)
(iv) Molality	Moles of solute per kg of solvent	$\frac{\text{mol}}{\text{kg}} = \text{mol kg}^{-1}$	(g)
(v) Pressure	Force per unit area	$\frac{\text{N}}{\text{m}^2} = \text{Pascal (Pa)}$	(c)

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Here are the common mistakes students make with this unit-matching question, along with the correct reasoning to avoid them.

Mistake 1: Confusing Molality and Molarity

This is the most frequent error. Students often swap the units for Molality and Molarity.

The Mistake: Matching Molality with $mol\ L^{-1}$ and Molarity with $mol\ kg^{-1}$ .
Why it happens: Both words sound similar and both involve "moles" of solute. The confusion lies in what the denominator represents.
How to Avoid:
- Molarity (M) = moles of solute per Liter of solution. The unit is $mol\ L^{-1}$ .
- Molality (m) = moles of solute per kilogram of solvent. The unit is $mol\ kg^{-1}$ .
- Memory Trick: Think of Molarity with a capital "L" (Liter) and Molality with a small "k" (kg). The "L" in Molarity reminds you of Liter.

Mistake 2: Forgetting that Mole Fraction is Unitless

Students often try to assign a unit like "mol" or a concentration unit to mole fraction.

The Mistake: Matching Mole fraction with (b) mol or (e) $mol\ L^{-1}$ .
Why it happens: Students see the word "mole" and assume it must have a unit related to moles.
How to Avoid:
- Definition: Mole fraction ( $x$ ) is the ratio of the number of moles of one component to the total number of moles of all components in the mixture.
- Key Insight: Since it is a ratio of two identical units (moles / moles), all units cancel out. The result is a pure number with no units.
- Correct Match: (ii) Mole fraction → (d) Unitless.

Mistake 3: Confusing Pressure with Surface Tension

The unit $Nm^{-1}$ (Newton per meter) is a trap. Students often associate "Newton" with force and incorrectly match it with Pressure.

The Mistake: Matching Pressure with (h) $Nm^{-1}$ .
Why it happens: Both Pressure and Surface Tension involve Force (Newtons) and length (meters), but in different ways.
How to Avoid:
- Pressure = Force per unit Area. Area is $m^2$ . So, unit is $N\ m^{-2}$ , which is also called a Pascal (Pa).
- Surface Tension = Force per unit Length. So, unit is $N\ m^{-1}$ .
- Correct Match: (v) Pressure → (c) Pascal. The unit $Nm^{-1}$ is not in the list for a correct match, so it is a distractor.

Mistake 4: Overlooking the Standard Unit for Mass

Students sometimes get confused because "kg" is listed as a separate option, while "g" appears inside other units. …

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