Define the following modes of expressing the concentration of a solution. Which of these modes are independent

NCERT Exemplar · Q55

Q. Define the following modes of expressing the concentration of a solution. Which of these modes are independent of temperature and why?

(i) w/w (mass percentage)

(ii) V/V (volume percentage)

(iii) w/V (mass by volume percentage)

(iv) ppm (parts per million)

(v) x (mole fraction)

(vi) M (Molarity)

(vii) m (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.

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

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: …

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Concentration modes based on mass (w/w, ppm, mole fraction, molality) are temperature-independent because mass does not change with temperature. Volume-based modes (V/V, w/V, molarity) change with temperature because volume expands or contracts.

The Core Idea: Mass vs. Volume

Temperature affects the volume of a solution — liquids expand when heated and contract when cooled. But mass? Mass stays constant regardless of temperature. So any concentration unit that uses mass in its definition will be temperature-independent, while any unit that uses volume will change when the temperature changes.

Let's examine each mode one by one.

1. w/w (Mass Percentage)

Mass percentage is defined as:

\text{w/w} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100

Both numerator and denominator are masses. Since mass does not change with temperature, w/w is temperature-independent.

2. V/V (Volume Percentage)

Volume percentage is:

\text{V/V} = \frac{\text{volume of solute}}{\text{volume of solution}} \times 100

Both volumes change with temperature (liquids expand/contract). So V/V is temperature-dependent.

Watch out

A common mistake is to think V/V is independent because both volumes change "in the same way." But they don't necessarily expand at the same rate — different liquids have different coefficients of thermal expansion. Even if they did, the ratio would still change because the volumes themselves change.

3. w/V (Mass by Volume Percentage)

This is:

\text{w/V} = \frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \times 100

The numerator (mass) is temperature-independent, but the denominator (volume) changes with temperature. So w/V is temperature-dependent.

4. ppm (Parts Per Million)

ppm can be expressed in different ways. The most common definition for solutions is:

\text{ppm} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6

Since this is a mass/mass ratio, ppm (when defined as mass/mass) is temperature-independent.

Tip

Be careful — ppm can also be defined as volume/volume (e.g., ppmV in gases). In that case it would be temperature-dependent. But in solution chemistry, ppm almost always means mass/mass.

5. x (Mole Fraction)

Mole fraction is:

x_A = \frac{n_A}{n_A + n_B + \dots}

where $n$ represents number of moles. Number of moles depends on mass and molar mass — both temperature-independent. So mole fraction is temperature-independent.

6. M (Molarity)

Molarity is:

$M = \frac{\text{moles of solute}}{\text{volume of solution in litres}}$ …

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Method: Temperature Dependence Analysis of Concentration Units

This method checks whether a concentration unit depends on volume (which changes with temperature) or only on mass/moles (which are temperature-independent).

Step-by-Step Reasoning

Identify which physical quantities change with temperature
- Mass → does not change with temperature
- Number of moles → does not change with temperature
- Volume → changes with temperature (expansion/contraction)
Classify each concentration unit

Mode	Definition	Depends on Volume?	Temperature Independent?
(i) w/w (mass percentage)	$\frac{\text{mass of solute}}{\text{mass of solution}} \times 100$	✗ No	✓ Yes
(ii) V/V (volume percentage)	$\frac{\text{volume of solute}}{\text{volume of solution}} \times 100$	✓ Yes	✗ No
(iii) w/V (mass by volume percentage)	$\frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \times 100$	✓ Yes	✗ No
(iv) ppm (parts per million)	$\frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6$	✗ No	✓ Yes
(v) x (mole fraction)	$\frac{\text{moles of solute}}{\text{total moles in solution}}$	✗ No	✓ Yes
(vi) M (Molarity)	$\frac{\text{moles of solute}}{\text{volume of solution (L)}}$	✓ Yes	✗ No

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Here are the common mistakes students make when tackling this exact question, along with the conceptual fixes to avoid them.

Mistake 1: Confusing which units are temperature-dependent

The Error: Students often say "Molarity is temperature dependent because it uses volume" but then incorrectly state that w/V (mass by volume percentage) is temperature independent.

Why it happens: They memorize "Molarity depends on temperature" but forget that any expression using volume (V) is temperature dependent because volume expands/contracts with heat.

How to avoid:

Rule of thumb: If the definition contains the word volume (liters, mL), it changes with temperature. If it contains only mass (grams, kg) or moles, it does not.
Correct classification:
- Temperature independent: w/w, ppm, mole fraction (x), molality (m) — all use mass or moles only.
- Temperature dependent: V/V, w/V, Molarity (M) — all use volume.

Mistake 2: Forgetting that ppm is just a scaled version of w/w

The Error: Students treat ppm as a completely separate concept and fail to realize it is mass/mass (or sometimes volume/volume) multiplied by $10^6$ .

Why it happens: The term "parts per million" sounds abstract, so students memorize it as a formula without linking it to w/w.

How to avoid:

Remember: ppm = $\frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6$ .
Since it is a ratio of masses, it is temperature independent for the same reason w/w is.

Mistake 3: Writing the formula for molality (m) incorrectly

The Error: Students write $m = \frac{\text{moles of solute}}{\text{volume of solvent in L}}$ (confusing it with molarity).

Why it happens: Both start with "m" and both involve moles. The difference is the denominator.

How to avoid:

Mnemonic: Molarity = Moles per Liter of solution (volume). molality = moles per kilogram of solvent (mass).
Key exam point: Molality uses mass of solvent (not solution), so it is temperature independent.

Mistake 4: Forgetting to specify "solvent" vs "solution" in definitions

The Error: In definitions, students write "mass of solute / mass of solution" for w/w but then write "moles of solute / mass of solution" for molality.

Why it happens: Careless reading of the denominator.

How to avoid:

Memorize the exact denominators:
- w/w: mass of solution
- m (molality): mass of solvent (in kg)
- M (molarity): volume of solution (in L)
- x (mole fraction): total moles of all components (solution)

Mistake 5: Not explaining why volume changes with temperature

The Error: Students say "Molarity depends on temperature because volume changes" but do not explain why volume changes.

Why it happens: They give the answer without the reasoning, losing marks in "explain" type questions.

How to avoid: …

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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 28 on this concept.

AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
$N = 6 \times 10^{23}\ mol^{-1}$
›Reveal solutionSolution
A straightforward percentage-composition → moles → atoms calculation; the answer is $1.2\times10^{21}$ Mg atoms.

Concept and Intuition

Percentage composition tells us the mass fraction of an element in a compound. Once we know the mass of that element, dividing by its atomic mass gives moles, and multiplying by Avogadro's number gives the actual atom count — the standard mass→mole→number chain.

Step-by-Step Solution

Mass of Mg in 2.0 g chlorophyll $= 2.4\% \times 2.0\ g = 0.024 \times 2.0 = 0.048$ g.

Moles of Mg $= \dfrac{0.048}{24} = 0.002$ mol. …
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AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
$g\,mol^{-1}$
›Reveal solutionSolution
Mustard gas, $(ClCH_2CH_2)_2S$ ( $C_4H_8Cl_2S$ , $M=159$ ), is 20.1% S and 44.65% Cl by mass.

Concept and Intuition

Percentage composition just needs the correct molecular formula and the atomic masses of the elements asked about, divided by the total molar mass.

Step-by-Step Solution

Mustard gas is bis(2-chloroethyl) sulfide, $(ClCH_2CH_2)_2S$ , i.e. $C_4H_8Cl_2S$ .

Molar mass check: $C_4=48$ , $H_8=8$ , $Cl_2=2\times35.5=71$ , $S=32$ . Sum $=48+8+71+32=159\,g\,mol^{-1}$ — matches the given value, confirming the formula.

Mass fraction of S $=\dfrac{32}{159}=0.2013 \Rightarrow 20.12\%$ (rounded as in the option).

Mass fraction of Cl $=\dfrac{71}{159}=0.4465 \Rightarrow 44.65\%$ .

So %S, %Cl (respectively) $=20.12,\ 44.65$ . …
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AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
$27°C$
›Reveal solutionSolution
Compute the mixture's average molar mass from the mass percentages, then apply the ideal-gas density formula $d=PM/RT$ to get $\approx 0.77\ \text{g L}^{-1}$ .

Concept and Intuition

For a gas mixture, the effective (average) molar mass is the total mass divided by total moles of all species combined. Once you have that average $M$ , the density of the mixture behaves just like a single ideal gas of that molar mass: $d = PM/RT$ .

Step-by-Step Solution

Assume 100 g of mixture: mass of $\mathrm{CH_4} = 64$ g, mass of $\mathrm{C_2H_6} = 36$ g.

Moles of $\mathrm{CH_4}$ ( $M=16$ g/mol) $= 64/16 = 4$ mol.

Moles of $\mathrm{C_2H_6}$ ( $M=30$ g/mol) $= 36/30 = 1.2$ mol.

Total moles $= 4+1.2 = 5.2$ mol; total mass $=100$ g.

Average molar mass $M_{avg} = 100/5.2 = 19.23$ g/mol.

Convert pressure: $P = 750/760 = 0.9868$ atm; $T = 27°C = 300$ K. …
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AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$\mathrm{H_2}$
›Reveal solutionSolution
Tests converting partial-pressure (mole fraction) data into a weight percentage using molar masses; the answer is 33.33%.

Concept and Intuition

Partial pressure is proportional to mole fraction (Dalton's law: $p_i = x_i P_{total}$ at constant T, V). Once we know the mole fractions of each gas, we can find the mass of each component (for an assumed total of 1 mole of gas mixture) using their molar masses, and hence the weight percentage.

Step-by-Step Solution

Mole fraction of $H_2$ : $x_{H_2} = \dfrac{p_{H_2}}{P_{total}} = \dfrac{1.778}{2} = 0.889$ .

Mole fraction of $O_2$ : $x_{O_2} = 1 - 0.889 = 0.111$ .

Assume 1 total mole of gas mixture. Mass of $H_2 = 0.889 \times 2\ \text{g/mol} = 1.778$ g. Mass of $O_2 = 0.111 \times 32\ \text{g/mol} = 3.552$ g.

Total mass = $1.778 + 3.552 = 5.33$ g. …
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AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$\mathrm{CaCO_3}$
›Reveal solutionSolution
Tests mass-balance on a two-component thermal decomposition mixture; solving simultaneous equations gives 54.34% $CaCO_3$ .

Concept and Intuition

Both carbonates decompose on strong heating: $CaCO_3 \to CaO + CO_2$ and $MgCO_3 \to MgO + CO_2$ , each losing $CO_2$ mass. Since the two salts lose mass in different fixed ratios (dictated by their molar masses), the residue mass depends on the mixture's composition — letting us solve for the unknown split.

Step-by-Step Solution

Let mass of $CaCO_3 = x$ g, so mass of $MgCO_3 = (1.84-x)$ g.

$CaCO_3$ (M = 100 g/mol) → $CaO$ (M = 56 g/mol): mass of CaO produced $= \dfrac{56}{100}x = 0.56x$ .

$MgCO_3$ (M = 84 g/mol) → $MgO$ (M = 40 g/mol): mass of MgO produced $= \dfrac{40}{84}(1.84-x)$ .

Total residue: $0.56x + \dfrac{40}{84}(1.84-x) = 0.96$ .

Compute $\dfrac{40}{84} = 0.4762$ : $0.56x + 0.4762(1.84) - 0.4762x = 0.96$ …
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AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$Q.The mole fractions of glucose and water in aqueous glucose solution are 0.0244 and 0.9756 respectively. What is the weight percentage (w/w) of glucose in this solution ? (A) 40 (B) 25 (C) 20 (D) 10$
›Reveal solutionSolution
Converting mole fraction to weight percentage for a glucose–water solution gives approximately 20% w/w glucose.

Concept and Intuition

Mole fraction tells us the mole ratio of components; to get weight percentage we must convert moles to mass using the molar masses (glucose $M = 180\ \mathrm{g/mol}$ , water $M = 18\ \mathrm{g/mol}$ ), then take the mass fraction.

Step-by-Step Solution

Assume a total of 1 mole of solution (mole fractions given directly as fractions of 1 mole total).

Moles of glucose $= 0.0244$ , moles of water $= 0.9756$ .

Mass of glucose $= 0.0244 \times 180 = 4.392\ \mathrm{g}$ .

Mass of water $= 0.9756 \times 18 = 17.5608\ \mathrm{g}$ .

Total mass of solution $= 4.392 + 17.5608 = 21.9528\ \mathrm{g}$ . …
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AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
$\mathrm{H_2}$
›Reveal solutionSolution
Converting the given weight percentage to moles (using molar masses 2 and 32 g/mol) gives a partial pressure ratio of $8:1$ .

Concept and Intuition

By Dalton's law of partial pressures, for a gas mixture at a common temperature and volume, the partial pressure of each component is proportional to its mole fraction (equivalently, to its number of moles). So converting mass percentages to moles (using each gas's molar mass) directly gives the pressure ratio.

Step-by-Step Solution

Take a convenient total mass, say $3\ \text{g}$ , so that $33.33\%=1/3$ works out to whole numbers: mass of H $_2$ $=1\ \text{g}$ , mass of O $_2$ $=2\ \text{g}$ .

Moles of H $_2$ : $n_{H_2} = \frac{1}{2} = 0.5\ \text{mol}$

Moles of O $_2$ : $n_{O_2} = \frac{2}{32} = 0.0625\ \text{mol}$ …
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AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
$Q.What are the mole fractions of glucose and water respectively, in 20% (w/w) aqueous glucose solution ? (C = 12 u; O = 16 u; H = 1 u) (A) 0.0244, 0.9756 (B) 0.04, 0.96 (C) 0.0636, 0.9364 (D) 0.0124, 0.9876$
›Reveal solutionSolution
Taking a 100 g basis of the 20% w/w solution gives 20 g glucose and 80 g water; converting to moles and dividing gives mole fractions 0.0244 (glucose) and 0.9756 (water).

Concept and Intuition

"20% w/w" means 20 g of solute per 100 g of solution, so the remaining 80 g is solvent. Mole fraction of each component is its moles divided by the total moles of all components.

Step-by-Step Solution

Basis: 100 g solution → 20 g glucose ( $\mathrm{C_6H_{12}O_6}$ , $M=180$ g/mol), 80 g water ( $M=18$ g/mol).

Moles of glucose $= 20/180 = 0.1111$ mol.

Moles of water $= 80/18 = 4.4444$ mol.

Total moles $= 0.1111 + 4.4444 = 4.5556$ mol.

Mole fraction of glucose $= 0.1111/4.5556 = 0.0244$ . …
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AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
$Q.In aqueous glucose solution, the mole fraction of water is 40 times to mole fraction of glucose. What is the weight percentage (w/w) of glucose in the solution? (A) 40 (B) 30 (C) 20 (D) 10$
›Reveal solutionSolution
This tests converting mole-fraction ratio into moles, then into a weight percentage. The answer is 20%.

Concept and Intuition

Mole fraction compares moles of a component to total moles, while weight percentage compares mass of a component to total mass. To go from one to the other we must first fix actual mole amounts (using the given ratio), then convert moles to grams using molar masses (glucose $M=180\ \mathrm{g\,mol^{-1}}$ , water $M=18\ \mathrm{g\,mol^{-1}}$ ).

Step-by-Step Solution

Let mole fraction of glucose $=x_2$ and of water $=x_1$ . Given $x_1 = 40x_2$ .

Since $x_1+x_2=1$ : $40x_2+x_2=1 \Rightarrow x_2=\dfrac{1}{41},\ x_1=\dfrac{40}{41}$ .

Take $n_2=1$ mol glucose $\Rightarrow n_1=40$ mol water (same ratio as mole fractions since total moles cancel).

Mass of glucose $=1\times180=180$ g. Mass of water $=40\times18=720$ g. …
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AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
$\mathrm{NaCl}$
›Reveal solutionSolution
This is a mixture stoichiometry problem: using the given %Na by mass to set up and solve a linear equation for the NaCl fraction, giving 45%.

Concept and Intuition

Each mole of NaCl and each mole of NaBr contributes exactly one mole of Na. So if we know the total mass of Na present, and express the unknown masses of NaCl and NaBr in terms of one variable, we can solve for the composition directly using molar masses.

Step-by-Step Solution

Total mixture mass = 4.0 g. Na is 30% of this by mass: mass of Na $= 0.30 \times 4.0 = 1.2$ g.

Moles of Na $= \dfrac{1.2}{23} = 0.052174$ mol.

Let mass of NaCl $= x$ g, so mass of NaBr $= (4.0 - x)$ g.

Molar mass of NaCl $= 23+35.5 = 58.5$ ; molar mass of NaBr $= 23+80=103$ .

Moles of Na from both salts: $\dfrac{x}{58.5} + \dfrac{4.0-x}{103} = 0.052174$ .

Multiply through by $58.5\times103 = 6025.5$ : $103x + 58.5(4.0-x) = 314.35$ . …
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AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
$\mathrm{NH_2CONH_2}$
›Reveal solutionSolution
Using the solution's density to get the total mass per litre, subtracting the urea mass gives the water mass; converting both to moles gives a urea mole fraction of about 0.038.

Concept and Intuition

Mole fraction requires the moles of each component in a defined amount of solution. Density lets us convert the given volume (1 dm³ = 1 L) into total solution mass, from which subtracting the known solute mass gives the solvent (water) mass.

Step-by-Step Solution

Total solution mass in 1 dm³ = density × volume = $1.02\ \mathrm{kg/dm^3}\times1\ \mathrm{dm^3} = 1020$ g.

Mass of urea = 120 g (given), so mass of water = $1020-120 = 900$ g.

Molar mass of urea, $\mathrm{NH_2CONH_2}$ : $2(14)+4(1)+12+16 = 28+4+12+16=60$ g/mol.

Moles of urea $= 120/60 = 2$ mol.

Moles of water $= 900/18 = 50$ mol. …
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AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
$_3$
›Reveal solutionSolution
Converting 1000 ppm of CaCO $_3$ (≈ 1 g/L for a dilute aqueous solution) into molarity using M = 100 g/mol gives $10^{-2}$ mol/L — option (D).

Concept and Intuition

"ppm" (parts per million) for a dilute aqueous solution is conventionally taken as mg of solute per litre of solution (since the solution's density is essentially that of water, 1 g/mL, at these very low concentrations). Once we have the mass concentration in g/L, dividing by the molar mass converts it to molarity directly.

Step-by-Step Solution

1000 ppm = 1000 mg solute per L of solution = 1 g/L (for a dilute aqueous solution where density ≈ 1 g/mL).

Molar mass of $\mathrm{CaCO_3}$ : $40 (\text{Ca}) + 12 (\text{C}) + 3\times16 (\text{O}) = 40+12+48 = 100$ g/mol. …
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Q.Define the following modes of expressing the concentration of a solution. Which of these modes are independent of temperature and why?

The Core Idea: Mass vs. Volume

1. w/w (Mass Percentage)

2. V/V (Volume Percentage)

3. w/V (Mass by Volume Percentage)

4. ppm (Parts Per Million)

5. x (Mole Fraction)

6. M (Molarity)

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