A solution contains \[1\] mole of water and \[4\] mole of ethanol. The mole fraction of water and ethanol will be
A. \[0.2\] Water + \[0.8\] ethanol
B. \[0.4\]water + \[0.6\] ethanol
C. \[0.6\]Water + \[0.8\] ethanol
D. \[0.8\] Water + \[0.2\] ethanol
Answer
288.3k+ views
Hint: Before we proceed into the problem, it is important to know the definition of mole fraction of a substance.
By dividing the component's mole by the sum of the moles of all the other components in the solution, the component's mole fraction can be computed. Calculating the component's moles requires dividing the component's stated mass by its molecular mass. By doing this we will get the desired answer of the mole fraction of water and ethanol.
Formula Used: The formula to determine the mole fraction is said to be
\[mole{\rm{ }}fraction{\rm{ }}of{\rm{ }}A{\rm{ }}(\;\chi A\;){\rm{ }} = \;\frac{{No.{\rm{ }}of{\rm{ }}moles{\rm{ }}of{\rm{ }}A}}{{Total{\rm{ }}number{\rm{ }}of{\rm{ }}moles}}\]
Complete Step-by-step answer: A solution contains \[1\] mole of water and \[4\] mole of ethanol which is given in the question.
The equation format of the given information is as below,
\[{X_{water}}\] = Water’s mole fraction = 1
\[{X_{ethanol}}\] = Ethanol’s mole fraction = 4
Now, we have to determine the mole fraction of a substance
Mole fraction’s formula is said to be as follows:
\[{X_A} = \frac{{{n_A}}}{{{n_A} + {n_B}}}\]---- (1)
First, let us predict the mole fraction of an ethanol
Therefore, on substituting the value on the above formula we get
\[{x_{ethanol}} = \frac{4}{{1 + 4}}\]
On further simplification, we have
\[{x_{ethanol}} = \frac{4}{5}\]
Now, we have to write the above fraction in decimal form, we get
\[{x_{ethanol}} = 0.8\]
Now, let us determine the mole fraction of water.
Therefore, on substituting the value on the mole fraction formula (1), we obtain
\[{x_{water}} = \frac{1}{{1 + 4}}\]
On further simplification, we get
\[{x_{water}} = \frac{1}{5}\]
Now, we have to write the above fraction in decimal form, we get
\[{x_{water}} = 0.2\]
Therefore, the mole fraction of water and ethanol will be \[0.2\] Water + \[0.8\] ethanol
Hence, the option A is correct
Note: Students shouldn’t get confused about Molality and molarity. Molality and molarity is two different concepts. While Molarity (M) is defined as the number of moles of a solute added to a solution per litre of solution, Molality (m) is defined as the number of moles of a solute per kilogram of a solvent. Students should also keep in mind that the mole fractions of each component in the solution added up should equal one.
By dividing the component's mole by the sum of the moles of all the other components in the solution, the component's mole fraction can be computed. Calculating the component's moles requires dividing the component's stated mass by its molecular mass. By doing this we will get the desired answer of the mole fraction of water and ethanol.
Formula Used: The formula to determine the mole fraction is said to be
\[mole{\rm{ }}fraction{\rm{ }}of{\rm{ }}A{\rm{ }}(\;\chi A\;){\rm{ }} = \;\frac{{No.{\rm{ }}of{\rm{ }}moles{\rm{ }}of{\rm{ }}A}}{{Total{\rm{ }}number{\rm{ }}of{\rm{ }}moles}}\]
Complete Step-by-step answer: A solution contains \[1\] mole of water and \[4\] mole of ethanol which is given in the question.
The equation format of the given information is as below,
\[{X_{water}}\] = Water’s mole fraction = 1
\[{X_{ethanol}}\] = Ethanol’s mole fraction = 4
Now, we have to determine the mole fraction of a substance
Mole fraction’s formula is said to be as follows:
\[{X_A} = \frac{{{n_A}}}{{{n_A} + {n_B}}}\]---- (1)
First, let us predict the mole fraction of an ethanol
Therefore, on substituting the value on the above formula we get
\[{x_{ethanol}} = \frac{4}{{1 + 4}}\]
On further simplification, we have
\[{x_{ethanol}} = \frac{4}{5}\]
Now, we have to write the above fraction in decimal form, we get
\[{x_{ethanol}} = 0.8\]
Now, let us determine the mole fraction of water.
Therefore, on substituting the value on the mole fraction formula (1), we obtain
\[{x_{water}} = \frac{1}{{1 + 4}}\]
On further simplification, we get
\[{x_{water}} = \frac{1}{5}\]
Now, we have to write the above fraction in decimal form, we get
\[{x_{water}} = 0.2\]
Therefore, the mole fraction of water and ethanol will be \[0.2\] Water + \[0.8\] ethanol
Hence, the option A is correct
Note: Students shouldn’t get confused about Molality and molarity. Molality and molarity is two different concepts. While Molarity (M) is defined as the number of moles of a solute added to a solution per litre of solution, Molality (m) is defined as the number of moles of a solute per kilogram of a solvent. Students should also keep in mind that the mole fractions of each component in the solution added up should equal one.
Recently Updated Pages
JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Maths Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 2) Chemistry Question Paper with Answer Key

Hydrogen and Its Type Important Concepts and Tips for JEE Exam Preparation

Concept of CP and CV of Gas - Important Concepts and Tips for JEE

[Awaiting the three content sources: Ask AI Response, Competitor 1 Content, and Competitor 2 Content. Please provide those to continue with the analysis and optimization.]

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

Understanding the Different Types of Solutions in Chemistry

Electron Gain Enthalpy and Electron Affinity Explained

Understanding Atomic Structure for Beginners

Other Pages
JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

NCERT Solutions For Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts Of Chemistry - 2026-27 Free PDF Download (Sign-in Required)

How to Convert a Galvanometer into an Ammeter or Voltmeter

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

Hybridisation in Chemistry – Concept, Types & Applications

