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# Importance of studying Mole Fraction

Last updated date: 02nd Aug 2024
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## Introduction to Mole Fraction

Mole Fraction describes the number of molecules contained within one component divided by the total number of molecules in a given mixture.  It is quite useful when two reactive-natured components are mixed together.  The ratio of both the components is known as the mole fraction.  It is a vital chapter of Class 11 that will also come for all the competitive exams that are held such as IIT ones, JEE Mains and NEET.

Understand the concept of mole fraction as it will help you get through the advanced concepts and principles of chemistry later. Use solved examples to find out how this concept has been used and learn to approach problems in the exercise judiciously.

It is advised that the students learn the concept of Mole Fraction primarily by practising problems and getting them verified against the answers on the articles provided on our website. This chapter is not to be left aside for last moment studying as this is an easy concept and can help in fetching good marks with little effort.

### Formula Related to the Mole Fraction

Consider a solution that consists of two substances A and B, then the mole fraction of each substance is:

Mole fraction of solute =  (Moles of Solute) / (Total number of moles of the solutes and the solvent).

= XA = (mol A) / ( mol A + mol B)

And

= XB = (mol B) / ( mol A + mol B)

Note:

1. In the given mixture, the sum of all the mole fractions present is equal to one

XA +  XB = 1.

1. When the mole fractions are multiplied by 100, they give the mole percentage.

2. Mole fraction is a unitless and dimensionless expression.

The image shows some chemical solutions of which we can find the mole fraction of its components.

The advantages of mole fraction are:

• Mole fraction has never been dependent on the temperature.

• To calculate the mole fraction, it is not necessary to find out the information about the density of the phase.

• The mole fraction is represented by the ratio of partial pressure to the total pressure of the mixture, in the case of an ideal gas mixture.

There is only one disadvantage of mole fraction, that is, mole fraction is not convenient for liquid solutions.

### Properties of Mole Fraction

The properties of mole fraction are:

• It is temperature independent which prevents it from variations when the temperature fluctuates. The knowledge of the densities of the phase involved is not necessary in the case of the mole fraction.

• By weighing off the actual masses of the constituents, a mixture of a known mole fraction can be prepared.

• The measure is symmetric, i.e., in the mole fraction x = 0.1 and x = 0.9, the roles of ‘solvent’ and ‘solute’ are reversed.

• In a ternary mixture, a mole fraction of a component can be expressed as functions of other components’ mole fraction and binary mole ratios.

Solved  Examples

Example 1:

Calculate the mole fraction of NaCl and H₂O, if 0.010 moles of NaCl is dissolved in 100 grams of pure water.

Solution:

The molecular weight of water is considered to be 18.0153 grams per mole.

Number of moles of water = 100 grams / 18.0153 grams = 5.56 moles

Mole fraction of NaCl = 0.100 moles / ( 5.56 moles + 0.10 moles)

= 0.100 moles / 5.66 moles

Mole fraction of NaCl = 0.018

Mole fraction of H₂O = 5.56 moles / 5.66 moles

Mole fraction of H₂O = 0.982

Example 2:

Calculate the mole fraction of acetone that is present in a solution which consists of 2 moles of benzene, 3 moles of carbon tetrachloride, and 5 moles acetone.

Solution:

Number of moles of acetone = 5 moles

Number of moles of carbon tetrachloride = 3 moles

Number of moles of benzene = 2 moles

Total number of moles in the solution = number of moles present in acetone + number of moles present in carbon tetrachloride + number of moles present in benzene

= 5 moles + 3 moles + 2 moles

= 10 moles

Mole fraction of acetone = (Number of moles in acetone) / (Total number of moles in the solution)

= 5 moles / 10 moles

= 0.5

Mole fraction of cabon tetrachloride = (Number of moles in carbon tetrachloride) / (Total number of moles in the solution)

= 3 moles / 10 moles

= 0.3

Mole fraction of benzene = (Number of moles in benzene) / (Total number of moles in the solution)

= 2 moles / 10 moles

= 0.2

Example 3:

A solution is prepared by mixing 30 grams of ethanol and 30 grams of water. Determine the mole fraction of each component.

Solution:

The molecular weight of ethanol is fixed at 46.07 g/mol.

The number of moles of ethanol present in the solution = 30 grams / 46.07 g/mol = 0.651 moles.

The molecular weight of water is fixed at 18g/mol.

The number of moles of water present in the solution =  30 grams / 18 g/mol = 1.67 moles

Total number moles = 0.651 + 1.67 =  2.321 moles

Mole fraction of ethanol = 0.651 moles / 2.321 moles = 0.28

Mole fraction of water = 1.67 moles / 2.321 moles = 0.72

## FAQs on Importance of studying Mole Fraction

1. How do students learn about the formulas related to Mole Fraction?

Students can learn about the formulae if they go to  Mole Fraction on our platform. This page is quite informative and has all the relevant details. Mole Fraction is an important chapter when it comes to Class 11 as well as for those who will need preparations for competitive exams later on. Understanding the concepts now and solidifying them will see to it that the student scores well during the tests. They need to revise this page before the tests too so that their concepts are clear and they are able to score well.

2. How do students perform well in their exams if Mole Fraction comes?

All students of Class 11, as well as, those looking to take engineering and other science-related competitive tests will need to be well-versed in this chapter. Mole Fraction is quite scoring if the formulae are understood and applied in the right manner. Going through the solved problems will help them to understand how each sum needs to be tackled. They can revise from Mole Fraction available on this page, as this will act as the ideal guidebook for them. Everything that they need to know has been included here. In fact, students can time themselves and then write down everything that they have learned from this page. This will increase their speed of writing and also shed light on how much they have picked up.

3. Where can students learn about the importance of Mole Fraction?

Students can learn about both if they refer to  Mole Fraction on our e-learning platform. Some advantages of mole fractions are that they have never been dependent on temperature, to calculate a mole fraction, it is not important to find out information about the density of the phase. If you go through this page on Mole Fraction, all your doubts will be taken care of in the best possible manner. Questions on their advantages and disadvantages have a probability of coming for tests.

4. Is Mole Fraction an important chapter for Class 11?

Mole Fraction is an important chapter for Class 11 as it is one of the fundamental chapters that you need to know about.  It is not difficult to score higher marks in this chapter as it is very logical. Going through the formulae and the solved examples will help the students understand the chapter in a thorough manner and also develop an interest in it.  Students can read from Mole Fraction on our platform to feel more confident in this chapter.

5. How can students clarify their doubts on Mole Fractions online?

All students who have doubts regarding the chapter on Mole Fractions can get them clarified by referring to our portal and then going to Mole Fraction, also available on this page.

This page is a comprehensive guide on Mole Fractions and is quite effective before the tests. Students can follow the content here so as to understand the concepts and write properly when they’re appearing for examinations. All doubts can be clarified if they scan this page properly and then make notes. This page is an ideal reference book for them to re-learn what they already know.