JEE Mains Chemistry · Some Basic Concepts of Chemistry

Molality, Mole Fraction and ppm

Molality, mole fraction, mass percent and parts per million, and converting between them through the density of the solution.

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Why this matters

Twenty-six PYQs, fourteen of them numerical and two from 2026. Thirteen convert between molarity, molality and mass percent through a density. Nine find a mole fraction or a mass percent, often of a mixture, and four use parts per million. Every one starts by fixing a basis: one litre, or one hundred grams, of solution.

Concept 1 of 3: Converting molarity, molality and mass percent

Density turns a volume of solution into a mass. Fix a basis, 1 L of solution or 100 g of it, find the masses of solute and solvent, and every concentration term follows.

Definition

Molality: $m=\frac{n_{\text{solute}}}{\text{mass of solvent (kg)}}$ .
Basis 1 L: mass of solution $=1000d$ g; mass of solvent $=1000d-nM_B$ . Here $d$ is the density in g mL $^{-1}$ , $M$ the molarity and $M_B$ the solute's molar mass.
From mass percent: $M=\frac{10\times(\%\text{ w/w})\times d}{M_B}$ , with $d$ in g mL $^{-1}$ .
v/v: turn each volume into a mass with its own density before anything else.

Molarity to molality

m=\frac{1000\,M}{1000\,d-M\,M_B}

Worked example

A 2 M solution of KCl (74.5 g mol

^{-1}

) has a density of

1.09

g mL

^{-1}

. Find its molality.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 26 June 2022 · Q31Moderate

Example 1 · Some Basic Concepts of Chemistry · Molality, Mole Fraction and ppm

A commercially sold conc.

HCl

35\% HCl

by mass. If the density of this commercial acid is 1.46

g/mL

, the molarity of this solution is: (Atomic mass:

Cl = 35.5amu,H = 1amu

)

Dividing by the solution's mass

Molality is per kilogram of solvent. For the 2 M KCl above, dividing by the whole 1.09 kg of solution gives 1.83 m; the right answer, 2.13 m, subtracts the 149 g of KCl first.

Drill 12 more on converting molarity, molality and mass percent

Concept 2 of 3: Mole fraction and mass percent of mixtures

Mole fraction is a share of the moles. Convert every mass to moles, then divide. When two solutions are mixed, add the solutes together and the solutions together first.

Definition

$x_A=\frac{n_A}{n_A+n_B+\ldots}$ , and all the fractions add to 1.
From molality in water: 1000 g of water is 55.5 mol, so $x=\frac{m}{m+55.5}$ .
Mass percent $=\frac{\text{mass of solute}}{\text{mass of solution}}\times100$ .
Mixing: total solute over total solution, for mass percent and mole fraction alike.

Mole fraction from molality

x_{\text{solute}}=\frac{m}{m+\frac{1000}{M_{\text{solvent}}}}

Worked example

Find the mole fraction of ethanol (

\mathrm{C_2H_5OH}

, 46 g mol

^{-1}

) in a solution of 23 g of ethanol in 36 g of water.

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The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 1 · Q29Moderate

Example 2 · Some Basic Concepts of Chemistry · Molality, Mole Fraction and ppm

What is the mole fraction of water in

10\%

by weight

(w/w)

of aqueous urea solution ? [Given : Molar mass of

H,O,C

and N are 1, 16, 12 and

14\text{ }g{\text{ }mol}^{- 1}

respectively]

Whose mole fraction?

Options usually carry both the solute's and the solvent's fraction, and the two add to 1. Read which one the question asks for.

Drill 8 more on mole fraction and mass percent of mixtures

Concept 3 of 3: Parts per million

ppm is a mass ratio multiplied by a million. In dilute water solutions, 1 ppm is 1 mg of solute per kilogram, which is 1 mg per litre.

Definition

$\text{ppm}=\frac{\text{mass of solute}}{\text{mass of solution}}\times10^{6}$ .
Dilute aqueous solutions: 1 ppm $=1$ mg kg $^{-1}\approx1$ mg L $^{-1}$ .
ppm of an element to mass of a compound: mass of the element, to moles, to moles of the compound (one Fe per $\mathrm{FeSO_4\cdot7H_2O}$ ), to mass.
A dense solution such as sea water: use its own density to find the mass of 1 L.

Parts per million

\text{ppm}=\frac{m_{\text{solute}}}{m_{\text{solution}}}\times10^{6}

Worked example

Drinking water contains 25 ppm of

\mathrm{Ca^{2+}}

. How many moles of

\mathrm{Ca^{2+}}

are there in 4 L (Ca

=40

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 4 April 2025 · Q48Moderate

Example 3 · Some Basic Concepts of Chemistry · Molality, Mole Fraction and ppm

Fortification of food with iron is done using

{FeSO}_{4}.7H_{2}O

. The mass in grams of the

{FeSO}_{4}.7H_{2}O

required to achieve 12 ppm of iron in 150 kg of wheat is (Nearest integer) [Given : Molar mass of Fe, S and O respectively are 56,32 and

16\text{ }g{\text{ }mol}^{- 1}

]

The element, not the compound

A limit in ppm of iron counts iron only. Find the moles of iron first, then the salt, whose molar mass is several times larger.

Drill 3 more on parts per million

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Converting molarity, molality and mass percent
Molarity to molality
$m=\frac{1000\,M}{1000\,d-M\,M_B}$
Mole fraction and mass percent of mixtures
Mole fraction from molality
$x_{\text{solute}}=\frac{m}{m+\frac{1000}{M_{\text{solvent}}}}$
Parts per million
Parts per million
$\text{ppm}=\frac{m_{\text{solute}}}{m_{\text{solution}}}\times10^{6}$

Watch out for (3)

Dividing by the solution's mass
→ Converting molarity, molality and mass percent
Whose mole fraction?
→ Mole fraction and mass percent of mixtures
The element, not the compound
→ Parts per million

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