
The mole fraction of the solute in a 1 molal aqueous solution is:
(a) 1
(b) 1.8
(c) 18
(d) 0.018
Answer
433.4k+ views
Hint: 1 molal aqueous solution means that 1 mole of a solute is dissolved in 1 Kg of water (the solvent). Now we can easily calculate the number of moles of the solvent and use that in order to find the mole fraction of the solute.
Complete step by step solution:
> First let us understand the terms mole fraction and molality.
> Mole fraction: The mole fraction of a particular component in a solution is the ratio of the number of moles of that component to the total number of moles in a solution. Therefore if we have a two component solution comprising of a solute and a solvent then:
$ mole\quad fraction\quad of\quad the\quad solute({ x }_{ 2 })=\cfrac { moles\quad of\quad the\quad solute({ n }_{ 2 }) }{ moles\quad of\quad solute({ n }_{ 2 })+moles\quad of\quad solvent({ n }_{ 1 }) } $
Similarly,
$ mole\quad fraction\quad of\quad the\quad solvent({ x }_{ 1 })=\cfrac { moles\quad of\quad the\quad solvent({ n }_{ 1 }) }{ moles\quad of\quad solute({ n }_{ 2 })+moles\quad of\quad solvent({ n }_{ 1 }) } $
> Also the summation of the mole fractions of all the components present in a solution is always unity.
$\sum { { x }_{ i }=1 } $
> Molality: It is used to denote the concentration of a solution. It refers to the ratio of the number of moles of solute to the mass of the solvent in kilogram. The SI unit of molality is molal (m).
$ Molality=\cfrac { moles\quad of\quad solute }{ Mass\quad of\quad solvent\quad in\quad Kg } $
> Now let us solve the question:
- It is given that the concentration of the aqueous solution is 1m. This means that 1 mole of the solute is dissolved in 1 Kg of water. The molar mass of water is 18 g/mol. Let us find out the number of moles of the solvent:
- The number of moles of water=$ \cfrac { Mass\quad of\quad water\quad in\quad g }{ Molar\quad mass\quad of\quad water } $
Therefore the moles of water present in the aqueous solution is= $ \cfrac { 1000g }{ 18g/mol } =55.56\quad mol$
Therefore the mole fraction of the solute is=$ \cfrac { 1\quad mol }{ 1\quad mol+55.56\quad mol } =0.018$
Therefore the correct answer is (d) 0.018.
Note: Always remember that mole fraction is a unit less quantity. Do not get confused between molality and molarity. While both of them are used to express the concentration of a solution but both of them are significantly different. Molarity is the ratio of the number of moles of the solute to the total volume of solution in litre while molality is the ratio of the moles of solute to the mass of the solvent in kilogram.
Complete step by step solution:
> First let us understand the terms mole fraction and molality.
> Mole fraction: The mole fraction of a particular component in a solution is the ratio of the number of moles of that component to the total number of moles in a solution. Therefore if we have a two component solution comprising of a solute and a solvent then:
$ mole\quad fraction\quad of\quad the\quad solute({ x }_{ 2 })=\cfrac { moles\quad of\quad the\quad solute({ n }_{ 2 }) }{ moles\quad of\quad solute({ n }_{ 2 })+moles\quad of\quad solvent({ n }_{ 1 }) } $
Similarly,
$ mole\quad fraction\quad of\quad the\quad solvent({ x }_{ 1 })=\cfrac { moles\quad of\quad the\quad solvent({ n }_{ 1 }) }{ moles\quad of\quad solute({ n }_{ 2 })+moles\quad of\quad solvent({ n }_{ 1 }) } $
> Also the summation of the mole fractions of all the components present in a solution is always unity.
$\sum { { x }_{ i }=1 } $
> Molality: It is used to denote the concentration of a solution. It refers to the ratio of the number of moles of solute to the mass of the solvent in kilogram. The SI unit of molality is molal (m).
$ Molality=\cfrac { moles\quad of\quad solute }{ Mass\quad of\quad solvent\quad in\quad Kg } $
> Now let us solve the question:
- It is given that the concentration of the aqueous solution is 1m. This means that 1 mole of the solute is dissolved in 1 Kg of water. The molar mass of water is 18 g/mol. Let us find out the number of moles of the solvent:
- The number of moles of water=$ \cfrac { Mass\quad of\quad water\quad in\quad g }{ Molar\quad mass\quad of\quad water } $
Therefore the moles of water present in the aqueous solution is= $ \cfrac { 1000g }{ 18g/mol } =55.56\quad mol$
Therefore the mole fraction of the solute is=$ \cfrac { 1\quad mol }{ 1\quad mol+55.56\quad mol } =0.018$
Therefore the correct answer is (d) 0.018.
Note: Always remember that mole fraction is a unit less quantity. Do not get confused between molality and molarity. While both of them are used to express the concentration of a solution but both of them are significantly different. Molarity is the ratio of the number of moles of the solute to the total volume of solution in litre while molality is the ratio of the moles of solute to the mass of the solvent in kilogram.
Recently Updated Pages
JEE Main 2025 Question Paper PDFs with Solutions Free Download

Difference Between Density and Volume: JEE Main 2024

Difference Between Series and Parallel Circuits: JEE Main 2024

Difference Between Analog and Digital: JEE Main 2024

Ammonium Hydroxide Formula - Chemical, Molecular Formula and Uses

Difference Between Work and Power: JEE Main 2024

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Electromagnetic radiation with maximum wavelength is class 11 chemistry JEE_Main

JEE Main 2025: Derivation of Equation of Trajectory in Physics

The correct order of electron affinity is A F Cl Br class 11 chemistry JEE_Main

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Types of Solutions

Other Pages
NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrocarbons

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics

Hydrocarbons Class 11 Notes: CBSE Chemistry Chapter 9

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry
