Which of the following units is useful in relating concentration of solution with its vapour pressure? (i) mol

NCERT Exemplar · Q1

Q. Which of the following units is useful in relating concentration of solution with its vapour pressure?

(i) mole fraction

(ii) parts per million

(iii) mass percentage

(iv) molality

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Concept understanding — Mass Percentage

Mass Percentage: The Intuition

Imagine you're making lemonade. You mix 50 grams of sugar into 200 grams of water. The total drink weighs 250 grams. Now, if someone asks, "How much of this drink is actually sugar?" — you're not just saying "50 grams." You want to say what fraction of the whole mixture is sugar, scaled to a convenient 100.

That's mass percentage. It answers: "Out of every 100 grams of the mixture, how many grams are this particular component?"

In our lemonade, sugar is 50 g out of 250 g total. That's $\frac{50}{250} = 0.2$ of the whole. Multiply by 100 to get the percentage: $0.2 \times 100 = 20\%$ . So, 20% of the drink's mass is sugar. If you had 100 g of this lemonade, 20 g of it would be sugar.

The Precise Definition

\text{Mass percentage of component} = \frac{\text{Mass of that component}}{\text{Total mass of mixture}} \times 100\%

The formula is simple, but the key is understanding what "total mass" means. It's the sum of masses of all components in the mixture — nothing more, nothing less.

Why It Matters in Chemistry

Mass percentage is one of the most common ways to express concentration — how much of a substance is present in a mixture. You'll see it in:

Solutions: "10% salt water" means 10 g of salt dissolved in enough water to make 100 g of solution (not 10 g salt + 100 g water — that would be 110 g total, giving only about 9.1%).
Alloys: "18-karat gold" is 75% gold by mass (18 parts gold out of 24 total parts).
Food labels: "Fat: 15%" means 15 g of fat per 100 g of the food.

Watch out

A Common Mistake

Students often think "10% salt solution" means 10 g salt + 100 g water. That's wrong. It means 10 g salt + 90 g water = 100 g total solution. The denominator is total mass, not the mass of the solvent alone.

Step-by-Step Example

Problem: A solution is made by dissolving 25 g of glucose in 175 g of water. Find the mass percentage of glucose.

Step 1: Identify the component you care about — glucose (25 g).

Step 2: Find the total mass of the mixture.

\text{Total mass} = 25\ \text{g (glucose)} + 175\ \text{g (water)} = 200\ \text{g}

Step 3: Apply the formula.

\text{Mass percentage of glucose} = \frac{25}{200} \times 100\% = 12.5\%

Interpretation: In every 100 g of this solution, 12.5 g is glucose and the rest (87.5 g) is water.

When to Use Mass Percentage vs. Other Measures

Mass percentage is ideal when:

You're working with solid mixtures or solutions where masses are easy to measure.
You want a concentration that doesn't change with temperature (unlike volume-based measures like molarity, which expand/contract with heat).

It's less useful when you need to count molecules (use mole fraction) or when volumes are more practical (use volume percentage).

One Final Check

If you ever get confused, go back to the lemonade. The question is always: "What fraction of the total weight is this one thing?" Multiply that fraction by 100, and you have your mass percentage.

Queries such as "mass percentage formula chemistry" and "mass percentage class 12 solutions" are common around this topic, which is a core concentration term introduced in the Solutions chapter of the NCERT/CBSE Class 12 Chemistry curriculum. Distinguishing it correctly from mass/volume percentage is a frequent numerical-question type in board exams and JEE Main.

Why this formula?

Let's break down Mass Percentage from first principles. The goal is to understand why the formula is what it is, not just to memorize it.

1. The Core Idea: "Part of a Whole"

Mass percentage answers a simple question: "If I break a mixture into 100 equal parts by mass, how many of those parts come from a specific component?"

Imagine you have a bowl of fruit salad. The total mass is 500 grams. The apples in it weigh 100 grams.

The apples are a part of the whole salad.
The whole salad is the total.

The mass percentage tells you the fraction of the total mass that is apples, but expressed "out of 100" (per cent).

2. The Natural First Step: The Fraction

Before we talk about "percentage," we talk about the fraction of the total:

\text{Fraction of component} = \frac{\text{Mass of component}}{\text{Total mass of mixture}}

For the apple example:

\frac{100 \text{ g}}{500 \text{ g}} = 0.2

This means 0.2 (or one-fifth) of the total mass is apples. This is the pure ratio — no scaling yet.

3. Why Multiply by 100?

A fraction like 0.2 is perfectly correct, but it's not intuitive for quick comparison. "Per cent" literally means "per hundred" (from Latin per centum).

To convert a fraction into a "per hundred" number, we multiply by 100:

\text{Percentage} = (\text{Fraction}) \times 100

So:

0.2 \times 100 = 20\%

This tells us: "Out of every 100 grams of fruit salad, 20 grams come from apples." That's much easier to visualize.

4. The Final Formula (The "Why" in One Line)

Putting the fraction and the "times 100" together gives the standard formula:

\text{Mass percentage} = \frac{\text{Mass of component}}{\text{Total mass of mixture}} \times 100\%

Why does this work?

Because it's just:

Find the proportion (part ÷ whole).
Scale that proportion to per hundred (× 100).

5. A Common Exam Trap (and Why It's Wrong)

Sometimes students write:

\text{Mass percentage} = \frac{\text{Mass of component}}{\text{Mass of solvent}} \times 100

This is incorrect. Why?

The denominator must be the total mass of the entire mixture (solute + solvent), not just the solvent.
The percentage tells you the share of the whole, not the share of one part relative to another.

Correct example:

10 g salt in 90 g water → total = 100 g.

Mass % of salt = $\frac{10}{100} \times 100 = 10\%$ (not $\frac{10}{90} \times 100 \approx 11.1\%$ ).

6. Quick Summary for Exams

Step	What to do	Why
1	Find the mass of the component	It's the "part"
2	Find the total mass of the mixture	It's the "whole"
3	Divide part by whole	Gives the fraction
4	Multiply by 100	Converts fraction to "per hundred"

Final takeaway: Mass percentage is just a scaled fraction — it makes comparisons easy by always using a base of 100.

The key idea is that Raoult’s law directly relates vapour pressure to the mole fraction of the solvent (or solute). Among the given options, only mole fraction appears in the law itself - the others are indirect or irrelevant. The correct answer is (i) mole fraction.

Why this question matters

When you study solutions and their colligative properties, the link between concentration and vapour pressure is fundamental. Raoult’s law states that the vapour pressure of a solvent above a solution equals the product of its mole fraction in the solution and its vapour pressure in the pure state:

P_{\text{solvent}} = x_{\text{solvent}} \cdot P_{\text{solvent}}^\circ

This is a direct, proportional relationship. The question asks which concentration unit is useful in relating concentration to vapour pressure - meaning which one appears naturally in the law itself.

Step-by-step reasoning

Recall Raoult’s law. For a solution of a non-volatile solute in a volatile solvent, $\frac{P^\circ - P}{P^\circ} = x_{\text{solute}}$ . No other concentration unit appears in this equation.
Examine each option
- (i) Mole fraction - dimensionless, appears directly in Raoult’s law. The natural variable for vapour-pressure relations.
- (ii) Parts per million (ppm) - a mass/volume-based ratio; does not appear in any vapour-pressure equation.
- (iii) Mass percentage - can be converted to mole fraction, but is not itself used in Raoult’s law.
- (iv) Molality - useful for boiling-point elevation and freezing-point depression, but not for vapour pressure.
Why the others are not “useful” in this context. Only mole fraction appears directly in the mathematical relationship; the others require conversion first.

Watch out

A common mistake is to pick molality because it is used for other colligative properties. But vapour-pressure lowering is directly proportional to mole fraction, not molality.

✓Final answer

The correct option is (i) mole fraction.

Method: Raoult's Law & Vapour Pressure Relation

The correct answer is (i) mole fraction.

Why Mole Fraction?

Vapour pressure of a solution is directly related to the mole fraction of the solvent via Raoult's Law:

P_{\text{solution}} = x_{\text{solvent}} \cdot P_{\text{solvent}}^\circ

where:

$P_{\text{solution}}$ = vapour pressure of the solution
$x_{\text{solvent}}$ = mole fraction of the solvent
$P_{\text{solvent}}^\circ$ = vapour pressure of pure solvent

Why Not the Others?

Unit	Reason it doesn't directly relate to vapour pressure
Parts per million (ppm)	Mass-based ratio; no direct link to mole fraction in Raoult's Law
Mass percentage	Also mass-based; doesn't appear in vapour pressure equations
Molality	Temperature-independent but still mass-based; not directly in Raoult's Law

Key Takeaway

Mole fraction is the only concentration unit that appears directly in Raoult's Law, making it the natural choice for relating concentration to vapour pressure.

\boxed{\text{(i) mole fraction}}

Correct Answer

(i) mole fraction

Why? Raoult's law states that the vapour pressure of a solution is directly proportional to the mole fraction of the solvent. Mathematically:

P_{\text{solution}} = X_{\text{solvent}} \cdot P_{\text{solvent}}^\circ

No other concentration unit appears directly in this law.

Common Mistakes & How to Avoid Them

1. Choosing mass percentage (option C)

The mistake: Students think "mass percentage" is the most common concentration unit, so it must relate to vapour pressure.

Why it's wrong: Mass percentage tells you grams of solute per 100 g of solution. Vapour pressure depends on the number of particles (moles) in the solution, not their mass. Two solutions with the same mass percentage can have very different vapour pressures if the solutes have different molar masses.

How to avoid: Always ask: "Does this unit count particles or just mass?" For vapour pressure, you need a particle-counting unit.

2. Choosing molality (option D)

The mistake: Students recall that molality is used in colligative properties (like boiling point elevation) and assume it works for vapour pressure too.

Why it's wrong: Molality ( $m$ ) is moles of solute per kg of solvent. While it is a particle-counting unit, Raoult's law uses mole fraction, not molality. Molality is useful for boiling point and freezing point, but not directly for vapour pressure.

How to avoid: Memorise the specific formula for each colligative property:

Vapour pressure → mole fraction
Boiling point elevation / freezing point depression → molality
Osmotic pressure → molarity

3. Choosing parts per million (option B)

The mistake: Students think "ppm is very precise, so it must be useful for vapour pressure."

Why it's wrong: ppm is just a scaled-up version of mass percentage (mg per kg). It still ignores particle count. It's used for trace concentrations (pollutants, minerals), not for vapour pressure calculations.

How to avoid: Remember that ppm, mass percentage, and volume percentage are all mass-based or volume-based units. Vapour pressure is a particle-based property.

4. Confusing mole fraction with mass fraction

The mistake: Students know "fraction" is involved, but they calculate mass fraction instead of mole fraction.

Example: For a solution of 10 g NaCl in 90 g water:

Mass fraction of NaCl = $10/100 = 0.1$
Mole fraction of NaCl = $\frac{10/58.5}{(10/58.5) + (90/18)} \approx 0.033$

These are very different — using mass fraction in Raoult's law gives a wrong answer.

How to avoid: Always convert given masses to moles before calculating mole fraction. Never substitute mass fraction directly.

Quick Summary Table

Unit	Counts particles?	Used in Raoult's law?
Mole fraction	✓ Yes	✓ Yes
Molality	✓ Yes	✗ No (used for $\Delta T_b$ , $\Delta T_f$ )
Mass percentage	✗ No	✗ No
ppm	✗ No	✗ No

Final tip: When you see "vapour pressure" in a question, immediately think mole fraction — it's the only unit that appears in Raoult's law directly.

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

GUJCET 2025Set 031 markMCQ
$Q.What will be mass percentage of aqueous solution of NaOH in which mole fraction of NaOH is 0.2? (A) 64.86% W/W (B) 35.71% W/W (C) 23.38% W/W (D) 27.78% W/W$
›Reveal solutionSolution
[!TLDR]

For a NaOH mole fraction of 0.2, the mass percentage of NaOH is about 35.71% (w/w).

Concept

Mole fraction gives the ratio of moles; converting to mass percentage requires multiplying moles by molar masses ( $M_{NaOH}=40$ , $M_{H_2O}=18$ g/mol).

Solution

Consider 1 mole of solution: $n_{NaOH}=0.2$ , $n_{H_2O}=1-0.2=0.8$ .

Mass of NaOH $=0.2\times40=8$ g.

Mass of water $=0.8\times18=14.4$ g.

Total mass $=8+14.4=22.4$ g.

Mass % of NaOH $=\dfrac{8}{22.4}\times100=35.71\%$ .

[!ANSWER]

(B) 35.71% W/W
GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$Q. If 22 gm of benzene (C6H6) dissolved in 122 gm of carbon tetrachloride (CCl4), calculate the mass percentage of benzene.$
(a) 84.72%
(b) 18.03%
(c) 15.28%
(d) 28.20%
›Reveal solutionSolution

Mass percentage = (mass of solute / total mass of solution) x 100.

Mass of benzene (solute) = 22 g

Mass of CCl4 (solvent) = 122 g

Total mass of solution = 22 + 122 = 144 g

Mass % of benzene = (22/144) x 100 = 15.28%

✓Final answer
$(c) 15.28%.$
GUJCET 2022Set 171 markMCQ
$NH_2CONH_2$
›Reveal solutionSolution

$x_{urea} = \dfrac{1}{1+55.55} \approx 0.01768$ .

Concept. 1 molal = 1 mol solute in 1000 g water.
$n_{water} = \frac{1000}{18} = 55.55 \text{ mol}$ $x_{urea} = \frac{1}{1 + 55.55} = \frac{1}{56.55} = 0.01768$
✓Final answer
$(B) 0.01768.$

ANSWER: (B)
GUJCET 2020Set 071 markMCQ
$Q.The molality of aqueous solution of any solute having mole fraction 0.25 is ______. (A) 33.33 m (B) 16.67 m (C) 18.52 m (D) 9.26 m$
›Reveal solutionSolution

$m=\dfrac{x_{\text{solute}}}{x_{\text{solvent}}\,M_{\text{solvent}}}=\dfrac{0.25}{0.75\times0.018}\approx18.52$ m.

Concept — mole fraction to molality (aqueous). Take 1 mol total: 0.25 mol solute, 0.75 mol water. Mass of water $=0.75\times18=13.5$ g $=0.0135$ kg.
$m=\frac{0.25}{0.0135}=18.52\ \text{mol kg}^{-1}.$
✓Final answer
$(C) 18.52 m$

ANSWER: (C)
GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQ
$Q. What is the weight to volume ppm of 0.05% w/v CaCl2 aqueous solution?$
(a) 0.05
(b) 500
(c) 50
(d) 5
›Reveal solutionSolution

ppm (parts per million) by weight/volume can be found directly by scaling the percentage w/v figure to a 10^6 basis.

0.05% w/v CaCl2 means 0.05 g of CaCl2 is present per 100 mL of solution.

Scaling to parts per million (parts per 10^6 by mass, taking 100 mL of dilute aqueous solution as approximately 100 g since density is close to 1 g/mL):

0.05 g per 100 g solution = (0.05/100) x 10^6 ppm = 0.0005 x 10^6 = 500 ppm.

(Equivalently: 0.05 g/100 mL = 0.5 g/L = 500 mg/L = 500 ppm, since 1 ppm = 1 mg/L for dilute aqueous solutions.)

✓Final answer
$(b) 500 ppm.$
GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQ
$Q. What is the concentration of solution in ppm when 5.0 x 10^-5 CO2 is dissolved in 100 ml solution.$
(a) 500
(b) 0.5
(c) 5
(d) 5.0 x 10^-5
›Reveal solutionSolution

Concentration in ppm = (mass solute / mass solution) x 10^6 = 0.5 ppm.

Parts per million: ppm = (mass of solute / mass of solution) x 10^6.

Given 5.0 x 10^-5 g of CO2 in 100 mL of dilute aqueous solution (density ~ 1 g/mL, so mass of solution ~ 100 g):

ppm = (5.0 x 10^-5 / 100) x 10^6 = 5.0 x 10^-7 x 10^6 = 0.5 ppm.

✓Final answer
$(b) 0.5.$
GUJCET 2015Set C1 markMCQ
$CH_3CH_2Cl$
›Reveal solutionSolution
[!TLDR] $0.1$ mol of chloroethane, 50% reacting $\Rightarrow 0.05$ mol $C_2H_4 = 1.4$ g.

Concept

Dehydrohalogenation ( $\beta$ -elimination) removes HCl from an alkyl halide to give an alkene: $CH_3CH_2Cl \xrightarrow{alc.\,KOH} CH_2\!=\!CH_2 + HCl$ (NCERT Haloalkanes).

Solution

Molar mass of $CH_3CH_2Cl = (2\times12) + (5\times1) + 35.5 = 64.5$ g/mol.

Moles taken $= \dfrac{6.45}{64.5} = 0.1$ mol.

Only 50% of the reagent reacts $\Rightarrow 0.05$ mol reacts.

$1$ mol $CH_3CH_2Cl$ gives $1$ mol ethene, so $0.05$ mol ethene forms.

Mass of $C_2H_4 = 0.05 \times 28 = 1.4$ g.

[!ANSWER] (A)

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Q.Which of the following units is useful in relating concentration of solution with its vapour pressure?

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Step-by-step reasoning

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