- Amal Augustine
- January 22, 2026
Master Mole Fraction with Solved Examples for NEET, JEE & CUET Aspirants
In chemistry,mole fraction with solved examples ,understanding how the components of a mixture are represented quantitatively is very important. In mole fraction with solved examples, one such concentration term is mole fraction, which is widely used in physical chemistry, especially in topics like solutions, colligative properties, Raoult’s law, and gaseous mixtures. Learning mole fraction with solved examples helps students gain clarity and confidence in numerical problems asked in school exams, NEET, JEE, and CUET.
What Is Mole Fraction?
The mole fraction of a component in a mixture is defined as the ratio of the number of moles of that component to the total number of moles of all components present in the mixture.
Mathematically,
Mole fraction of A(XA)=moles of Atotal moles of all components\text{Mole fraction of A} (X_A) = \frac{\text{moles of A}}{\text{total moles of all components}}
Mole fraction has no unit because it is a ratio. The sum of mole fractions of all components in a mixture is always equal to 1.
Why Mole Fraction Is Important
Mole fraction is independent of temperature and pressure, making it a very reliable concentration term. It is particularly useful in:
-
Vapour pressure calculations
-
Raoult’s law
-
Henry’s law
-
Colligative properties
-
Gas mixtures
This is why mole fraction with solved examples is a high-scoring and frequently tested concept in competitive examinations.
In Mole Fraction with Solved Example 1: Mole Fraction in a Liquid Solution
Problem:
8 g of NaOH is dissolved in 18 g of water. Calculate the mole fraction of NaOH.
Solution:
Molar mass of NaOH = 40 g/mol
Moles of NaOH = 8 ÷ 40 = 0.2 mol
Molar mass of water = 18 g/mol
Moles of water = 18 ÷ 18 = 1 mol
Total moles = 0.2 + 1 = 1.2 mol
XNaOH=0.21.2=0.167X_{\text{NaOH}} = \frac{0.2}{1.2} = 0.167
This example clearly shows how to approach mole fraction with solved examples step by step.
In Mole Fraction with Solved Examples 2: Mole Fraction in an Aqueous Solution
Problem:
Calculate the mole fraction of solute in a 1.00 molal aqueous solution.
Solution:
1 molal means 1 mole of solute is dissolved in 1000 g (1 kg) of water.
Moles of water = 1000 ÷ 18 = 55.5 mol
Total moles = 1 + 55.5 = 56.5
Xsolute=156.5=0.0177X_{\text{solute}} = \frac{1}{56.5} = 0.0177
This is a very common numerical type when studying mole fraction with solved examples in colligative properties.
In Mole Fraction with Solved Examples 3: Mole Fraction in Gas Mixtures
Problem:
A mixture contains 2 moles of oxygen and 6 moles of nitrogen. Find the mole fraction of oxygen.
Solution:
Total moles = 2 + 6 = 8
XO2=28=0.25X_{O_2} = \frac{2}{8} = 0.25
Gas-phase problems like this are often included while teaching mole fraction with solved examples.
Key Properties of Mole Fraction
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It is dimensionless (no unit)
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Independent of temperature
-
Sum of mole fractions = 1
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Suitable for liquid and gaseous solutions
Because of these advantages, mole fraction is preferred over molarity in many thermodynamic calculations.
Common Mistakes to Avoid
While solving mole fraction with solved examples, students often:
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Forget to convert mass into moles
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Use volume instead of moles
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Forget to include all components in total moles
Careful reading and step-by-step calculation can easily avoid these errors.
MCQs on Mole Fraction with Solved Examples:
1.
The mole fraction of a solvent in an aqueous solution is 0.8. The molality (mol kg⁻¹) of the solution is:
a) 13.88 × 10⁻²
b) 13.88 × 10⁻³
c) 13.88
d) 13.88 × 10⁻¹
Answer: c
2.
A solution of sodium sulphate contains 92 g of Na⁺ ions per kg of water. The molality of Na⁺ ions is:
a) 4
b) 16
c) 132
d) 8
Answer: a
3.
The amount of sugar (C₁₂H₂₂O₁₁) required to prepare 2 L of 0.1 M aqueous solution is:
a) 17.1 g
b) 68.4 g
c) 136.8 g
d) 34.2 g
Answer: b
4.
8 g NaOH dissolved in 18 g H₂O. Mole fraction of NaOH and molality respectively are:
a) 0.2, 11.11
b) 0.167, 22.20
c) 0.2, 22.20
d) 0.167, 11.11
Answer: d
5.
10 mmol Ca(OH)₂ and 2 g Na₂SO₄ in 100 mL. Mass of CaSO₄ formed and [OH⁻] are:
a) 13.6 g, 0.28 M
b) 1.9 g, 0.28 M
c) 13.6 g, 0.14 M
d) 1.9 g, 0.14 M
Answer: b
6.
Strength of 11.2 volume H₂O₂ solution is:
a) 1.7 %
b) 34 %
c) 13.6 %
d) 3.4 %
Answer: d
7.
On mixing 20 mL acetone with 60 mL chloroform, total volume is:
a) Less than 80 mL
b) More than 80 mL
c) Exactly 80 mL
d) Unpredictable
Answer: a
8.
62 g ethylene glycol in 250 g water cooled to −10°C. Ice separated (g) is:
a) 64
b) 48
c) 34
d) 16
Answer: a
9.
Ideal solution at 350 K with vapour pressures 7×10³ Pa and 12×10³ Pa. Vapour composition for 40% A is:
a) XA = 0.76, XB = 0.24
b) XA = 0.28, XB = 0.72
c) XA = 0.40, XB = 0.60
d) XA = 0.37, XB = 0.63
Answer: b
10.
Mole fraction of solute in 1.00 molal aqueous solution is:
a) 0.177
b) 0.0344
c) 1
d) 0.0177
Answer: d
11.
25 mL HCl needs 30 mL of 0.1 M Na₂CO₃. Volume of HCl needed to titrate 30 mL of 0.2 M NaOH is:
a) 75 mL
b) 25 mL
c) 12.5 mL
d) 50 mL
Answer: b
12.
Example of an ideal solution is:
a) CCl₄ + CHCl₃
b) CCl₄ + Toluene
c) Toluene + Benzene
d) Benzene + Acetone
Answer: c
13.
Mole fraction of gas dissolved in solution is:
a) Proportional to partial pressure of gas above solution
b) Proportional to partial pressure in solution
c) Proportional to square of pressure above solution
d) Proportional to square of pressure in solution
Answer: a
14.
Molarity of urea solution by dissolving 15 g urea in 500 cm³ water is:
a) 2 M
b) 0.5 M
c) 0.125 M
d) 0.005 M
Answer: b
15.
1 mol NaCl dissolved in 1 kg water. Boiling point at 1.013 bar is:
a) 373.15 K
b) 373.67 K
c) 374.19 K
d) 375 K
Answer: b
16.
If attraction A–B > A–A and B–B, solution shows:
a) Positive deviation
b) Negative deviation
c) No deviation
d) Cannot predict
Answer: b
17.
Density of sulphuric acid used in lead accumulator is:
a) 1.8 g mL⁻¹
b) 1.2 g mL⁻¹
c) 1.5 g mL⁻¹
d) 2.0 g mL⁻¹
Answer: b
18.
Molarity of solution of 30 g Co(NO₃)₂·6H₂O in 4.3 L is:
a) 0.023 M
b) 0.23 M
c) 0.046 M
d) 0.46 M
Answer: a
19.
100 mL flask with H₂ at 200 torr and 200 mL flask with He at 100 torr are connected. Total pressure is:
a) 104 torr
b) 163.33 torr
c) 279 torr
d) 133.33 torr
Answer: d
20.
Boiling point of solution of acenaphthalene in CHCl₃ is:
a) 61.9°C
b) 62°C
c) 52.2°C
d) 62.67°C
Answer: d
21.
Molality of 720 g pure water is:
a) 40
b) 4
c) 55.5
d) None
Answer: c
22.
Correct statement:
a) Surface tension decreases with temperature
b) Vapour pressure decreases with temperature
c) Viscosity decreases with decrease in temperature
d) Droplets flatten in zero gravity
Answer: a
23.
Equimolar urea and KCl solutions separated by membrane. Result:
a) No net flow
b) Solvent flows from KCl to urea
c) Solvent flows from urea to KCl
d) Cannot predict
Answer: c
24.
Normality of salt solution after neutralisation and dilution is:
a) 2(x+y)xyN\frac{2(x+y)}{xy}N
b) xy2(x+y)N\frac{xy}{2(x+y)}N
c) 2xyx+yN\frac{2xy}{x+y}N
d) x+yxyN\frac{x+y}{xy}N
Answer: b
25.
Which solution turns violet on adding lime juice?
a) NaI
b) KI + NaIO₃
c) NaI + KI
d) KIO₃ + NaIO₃
Answer: b
26.
Molality of solution with 15.20 g urea in 150 g water is:
a) 1.689 m
b) 0.1689 m
c) 0.5922 m
d) 0.2533 m
Answer: a
27.
Which depends on temperature?
a) Molarity
b) Mole fraction
c) Weight percentage
d) Molality
Answer: a
28.
Volume of O₂ liberated at STP from 15 mL of 20 volume H₂O₂ is:
a) 100 mL
b) 150 mL
c) 200 mL
d) 300 mL
Answer: d
29.
Molality of 3 M methanol solution (density = 0.9 g cm⁻³) is:
a) 3.73
b) 3
c) 3.33
d) 3.2
Answer: a
30.
Volume of water to be added to 100 cm³ of 0.5 N H₂SO₄ to get decinormal solution is:
a) 100 cm³
b) 450 cm³
c) 500 cm³
d) 400 cm³
Answer: d
Conclusion on Mole Fraction with Solved Examples
Understanding mole fraction with solved examples is essential for mastering solution chemistry. In mole Fraction with Solved Examples, it provides a clear picture of composition, works well across temperatures, and plays a central role in vapour pressure and colligative property calculations. With regular practice of mole Fraction with Solved examples and proper understanding of mole conversion, aspirsants can solve mole-fraction numericals accurately and confidently.
Amal Augustine is the founder of ExQuizMe, a dynamic learning and quiz platform built to make education engaging, competitive, and fun. A passionate learner and an academic achiever, Amal completed his schooling at Government HSS Manjapra, graduating with 92.5% in Computer Science. He later earned his degree from St. Stephen’s College, University of Delhi, one of India’s most prestigious arts and science institutions.
Currently, Amal is pursuing his Master’s degree at National Sun Yat-sen University, Taiwan, where he continues to deepen his interest in research and technology. Throughout his school and college years, he won 50+ national-level interschool and collegiate quiz competitions, was
Beyond academics, Amal Augustine is an avid reader of science journals, a dedicated research student, and a technology enthusiast who loves programming and exploring the world of Computer Science. Through ExQuizMe, he aims to make learning accessible, enjoyable, and empowering for students across the globe.