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.

COMEDK 2026Set 2026-M1 markMCQ
$X$
›Reveal solutionSolution
Mass percent is the mass of solute divided by the total mass of solution, times 100. Here, the solute mass is 4.0 g, and the solvent (water) mass is 2.0 moles × 18 g/mol = 36 g, so total mass = 40 g, giving mass percent = (4/40)×100 = 10%. The correct option is (D).

Concept & Intuition

Mass percent tells you how many grams of solute are present in every 100 grams of solution. It’s a simple ratio:
$\text{mass percent} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100\%$
The trick here is that the solvent (water) is given in moles, not grams. So the first step is always to convert moles of water to grams using its molar mass (18 g/mol). Once everything is in grams, the calculation is straightforward.

Step-by-step solution

Find the mass of water (solvent) We have 2.0 moles of water. The molar mass of water is $18.0\ \text{g/mol}$ .

$\text{mass of water} = 2.0\ \text{mol} \times 18.0\ \frac{\text{g}}{\text{mol}} = 36\ \text{g}$

Find the total mass of the solution The solution contains the solute (4.0 g of X) plus the solvent (36 g of water).

$\text{total mass} = 4.0\ \text{g} + 36\ \text{g} = 40\ \text{g}$

Calculate the mass percent

$\text{mass percent of X} = \frac{4.0\ \text{g}}{40\ \text{g}} \times 100\% = 0.10 \times 100\% = 10\%$
Watch out

A common mistake is to forget to convert moles of water to grams, and instead use 2.0 as if it were grams. That would give $\frac{4}{4+2} \times 100 \approx 67\%$ , which isn’t even among the options — but it’s a trap to watch for.

Tip

Always check units: if the solvent is given in moles, convert to grams first. The molar mass of water (18 g/mol) is a constant you should know by heart for such problems.

✓Final answer
$The correct option is (D) . ANSWER: D$
KCET 2025Set D-41 markMCQ
$Q.Which of the following methods of expressing concentration are unitless? (A) Mole fraction and Mass percent (W/W) (B) Molality and Mole fraction (C) Mass percent (W/W) and Molality (D) Molality and Molarity$
›Reveal solutionSolution
A concentration term is unitless only when it is a ratio of two quantities of the same kind — mole/mole or mass/mass — so the units cancel.

Step 1 — Write each concentration measure with its units.

Mole fraction

$x_A = \frac{n_A}{n_A + n_B} = \frac{\mathrm{mol}}{\mathrm{mol}}$
Moles divided by moles $\Rightarrow$ unitless (and it always lies between 0 and 1).

Mass percent (W/W)

$\%\,(w/w) = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100 = \frac{\mathrm{g}}{\mathrm{g}} \times 100$
Grams divided by grams $\Rightarrow$ unitless (the "%" is a pure number, not a unit).

Molality

$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \mathrm{mol\,kg^{-1}}$
Moles divided by mass — two different kinds of quantity $\Rightarrow$ has units.

Molarity

$M = \frac{\text{moles of solute}}{\text{volume of solution in L}} = \mathrm{mol\,L^{-1}}$
Moles divided by volume $\Rightarrow$ has units.

Step 2 — Apply the test.

The unitless pair is therefore mole fraction and mass percent (W/W).

Step 3 — Eliminate.

(B) Molality has units ( $\mathrm{mol\,kg^{-1}}$ ) — fails.

(C) Molality has units — fails.

(D) Both molality and molarity have units — fails outright.

Only (A) lists two genuinely dimensionless measures.

Aside worth knowing: the fact that mole fraction, molality and mass percent are all temperature-independent (they involve no volume, which expands on heating) — whereas molarity is temperature-dependent — is a closely related and frequently examined point. But the specific question here is about units, and on that test molality is excluded while mass percent is included.

✓Final answer
$The correct option is (A) — Mole fraction and Mass percent (W/W).$

ANSWER: A
COMEDK 2025Set 2025-M1 markMCQ
$\mathrm{Cl}=35.5 \mathrm{~g} / \mathrm{mol}$
›Reveal solutionSolution
The key is to use the given chlorine mass percentage to find the molar mass of the haloalkane, then compute the number of molecules in 0.1 g, and finally multiply by one Cl atom per molecule. The result is about $9.328 \times 10^{20}$ Cl atoms, so the correct option is (C).

Concept & Intuition

We have a haloalkane (an alkane with one chlorine atom replacing a hydrogen). The problem tells us that chlorine makes up 55% of the mass of one molecule. That means if we know the mass of one mole of the compound, we can find how many moles of Cl are in a sample. Since each molecule has exactly one Cl atom, the number of Cl atoms equals the number of molecules. So the plan: find the molar mass from the percentage, then convert 0.1 g to moles, then to atoms via Avogadro’s number.

Step-by-step solution

Relate percentage to molar mass Let $M$ be the molar mass of the haloalkane (in g/mol). One mole of the compound contains one mole of chlorine atoms, which has mass $35.5$ g. The percentage by mass of chlorine is:

$\frac{35.5}{M} \times 100\% = 55\%$
So:
$\frac{35.5}{M} = 0.55$
Solving:
$M = \frac{35.5}{0.55} = 64.545\ldots \ \text{g/mol}$
(We can keep it as $\frac{35.5}{0.55}$ for now.)

Find moles of haloalkane in 0.1 g Moles of compound:

$n = \frac{\text{mass}}{M} = \frac{0.1}{35.5/0.55} = \frac{0.1 \times 0.55}{35.5}$
Simplify:
$n = \frac{0.055}{35.5} \ \text{mol}$

Number of molecules (and thus Cl atoms) Since each molecule has one Cl atom, the number of Cl atoms is:

$N = n \times N_A = \frac{0.055}{35.5} \times 6.022 \times 10^{23}$
Compute step by step:
$\frac{0.055}{35.5} = \frac{55}{35500} = \frac{11}{7100} \approx 0.0015493$
Multiply by Avogadro’s number:
$N \approx 0.0015493 \times 6.022 \times 10^{23} = 9.328 \times 10^{20}$

Match with options The value $9.328 \times 10^{20}$ matches option (C) exactly (option D is off by a factor of 1000, a common mistake if you forget to convert grams to moles properly).

Watch out

A classic pitfall is to forget that the percentage refers to mass, not moles. Another is to compute the number of Cl atoms as if there were multiple Cl atoms per molecule — but the problem says “a single Chlorine atom per molecule.”

Tip

You can also think: 55% of the mass is Cl, so in 0.1 g sample, mass of Cl = $0.055$ g. Then moles of Cl = $0.055 / 35.5$ , and atoms = that times Avogadro’s number — same calculation, even faster.

✓Final answer
$The correct option is (C) . ANSWER: C$
KCET 2024Set B-21 markMCQ
$Q. For which one of the following mixtures is composition uniform throughout? (A) Sand and water (B) Grains and pulses with stone (C) Mixture of oil and water (D) Dilute aqueous solution of sugar$
›Reveal solutionSolution
"Uniform composition throughout" is the definition of a homogeneous mixture (a true solution) — only the sugar solution qualifies.

Step 1 — The concept.

Mixtures are classified by whether their composition is the same at every point:

Homogeneous mixture (solution): solute particles are of molecular/ionic size ( $< 1\,$ nm), uniformly dispersed. Every sample drawn from anywhere has the same composition. Only one phase is visible.

Heterogeneous mixture: two or more distinguishable phases; composition varies from point to point.

Step 2 — Test each option.

(A) Sand and water — sand particles are large and insoluble; they settle to the bottom. Two visible phases → heterogeneous ✗

(B) Grains and pulses with stone — plainly separable solids, each retaining its identity; a scoop from one corner differs from another → heterogeneous ✗

(C) Mixture of oil and water — oil is non-polar, water is polar; they are immiscible and separate into two layers with a clear meniscus → heterogeneous ✗

(D) Dilute aqueous solution of sugar — sucrose is polar and hydrogen-bonds with water, so it dissolves completely into individual molecules dispersed at random. Every drop tastes equally sweet and has the same concentration → homogeneous ✓

Step 3 — Commit. Only (D) has uniform composition throughout.

✓Final answer
$The correct option is (D) — Dilute aqueous solution of sugar.$

ANSWER: D
KCET 2022Set B-31 markMCQ
$18g$
›Reveal solutionSolution
Mole fraction is the ratio of moles of one component to total moles. Here, water’s mole fraction is $0.1$ , so the correct option is (C).

The concept here is mole fraction — a way to express concentration in terms of the number of particles (moles) rather than mass. In a mixture, the mole fraction of a component is simply the number of moles of that component divided by the total number of moles of all components. It’s dimensionless and always lies between 0 and 1.

Why does this matter? Because mole fraction directly relates to partial pressures in gases and colligative properties in solutions. For this problem, we just need to convert the given masses into moles using molar masses, then compute the ratio.

Find the moles of water. Water ( $H_2O$ ) has a molar mass of $18 \, \text{g/mol}$ . Given $18 \, \text{g}$ of water:

$n_{\text{water}} = \frac{18}{18} = 1 \, \text{mol}.$

Find the moles of ethyl alcohol. Ethyl alcohol ( $C_2H_5OH$ ) has a molar mass of $46 \, \text{g/mol}$ (carbon: $2 \times 12 = 24$ , hydrogen: $6 \times 1 = 6$ , oxygen: $16$ , total $24+6+16=46$ ). Given $414 \, \text{g}$ of alcohol:

$n_{\text{alcohol}} = \frac{414}{46} = 9 \, \text{mol}.$

Calculate total moles.

$n_{\text{total}} = n_{\text{water}} + n_{\text{alcohol}} = 1 + 9 = 10 \, \text{mol}.$

Compute mole fraction of water.

$x_{\text{water}} = \frac{n_{\text{water}}}{n_{\text{total}}} = \frac{1}{10} = 0.1.$
Watch out

A common mistake is to use masses directly instead of converting to moles. Mass ratio ( $18/432 = 0.0417$ ) is not the mole fraction — always divide by molar mass first.

Tip

Notice that $414$ is $9 \times 46$ , so the alcohol moles come out cleanly. In exam problems, numbers are often chosen to give neat results — trust the arithmetic.

✓Final answer
$0.1$
KCET 2022Set B-31 markMCQ
$Q. Vacant space in body centered cubic lattice unit cell is about (A) 23% (B) 46% (C) 32% (D) 10%$
›Reveal solutionSolution
Vacant space $= 100\% -$ packing efficiency; for bcc the packing efficiency is $68\%$ , so $32\%$ is empty.

Step 1 — Set up the bcc geometry.

A bcc unit cell has:

$8$ corner atoms, each shared by $8$ cells $\Rightarrow 8\times\tfrac18 = 1$ atom

$1$ atom fully inside at the body centre $\Rightarrow 1$ atom

$Z = 2\ \text{atoms per unit cell}$
Step 2 — Relate radius to edge length.

In bcc the atoms touch along the body diagonal, not along the edge. The body diagonal of a cube of edge $a$ has length $\sqrt3\,a$ , and it contains $4$ radii (corner atom radius + full central atom + corner atom radius):
$\sqrt{3}\,a = 4r \quad\Longrightarrow\quad r = \frac{\sqrt3\,a}{4}$
Step 3 — Compute the packing efficiency.
$\text{P.E.} = \frac{Z \times \tfrac43\pi r^3}{a^3} = \frac{2 \times \tfrac43\pi\left(\tfrac{\sqrt3 a}{4}\right)^3}{a^3}$ $= \frac{\tfrac83\pi \cdot \tfrac{3\sqrt3 a^3}{64}}{a^3} = \frac{\sqrt3\,\pi}{8} = \frac{1.732 \times 3.1416}{8} \approx 0.680$
So the atoms fill $68\%$ of the cell.

Step 4 — Vacant space.
$\text{Vacant} = 100\% - 68\% = \boxed{32\%}$
(For comparison: fcc/ccp is $74\%$ filled → $26\%$ vacant; simple cubic is $52.4\%$ filled → $47.6\%$ vacant. Option (B) $46\%$ is the trap meant for simple cubic, and (A) $23\%$ echoes fcc.)

✓Final answer
$The correct option is (C) — 32%.$

ANSWER: C

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

Why this question matters

Step-by-step reasoning

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