Numerical Problems on Mole Fraction

Physics Chemistry  Biology  Mathematics
Science > Chemistry > Solutions and Colligative PropertiesYou are Here

Example – 01:

  • 23 g of ethyl alcohol (molar mass 46 g mol-1) is dissolved in 54 g of water (molar mass 18 g mol-1). Calculate the mole fraction of ethyl alcohol and water in solution.
  • Given: Mass of solute = WB = 23 g, Molar mass of solute = MB = 46 g mol-1,  Mass of solvent = WA = 54 g, Molar mass of solvent = MA = 18 g mol-1,
  • To Find: Mole fractions xB =? xA = ?
  • Solution:

Number of moles of solute (ethyl alcohol) = nB = 23 g/ 46 g mol-1 = 0.5 mol

Number of moles of solvent (water) = nA = 54 g/ 18 g mol-1 = 3 mol

Total number of moles = n+ n= 0.5 + 3 = 3.5 mol

Mole fraction of solute (ethyl alcohol) = xB = nB/(n+ nB) = 0.5/3.5 = 0.1429



Mole fraction of solvent (water) = xA = nA/(n+ nB) = 3/3.5 = 0.8571

Example – 02:

  • 4.6 cm3 of methyl alcohol is dissolved in 25.2 g of water. Calculate a) percentage by mass of methyl alcohol b) mole fraction of methyl alcohol and water. Given density of methyl alcohol = 0.7952 g cm-3, and C = 12, H = 1 and O = 16.
  • Given: Volume of solute (methyl alcohol) = 4.6 cm3, mass of solvent (water) = 25.2 g,  Density of methyl alcohol = d = 0.7952 g cm-3,
  • To Find: percentage by mass of methyl alcohol = ?,  mole fraction of methyl alcohol and water = ?
  • Solution:

Mass of methyl alcohol = Volume x density = 4.6 cm3 x 0.7952 g cm-3 = 3.658 g

Mass of solution = Mass of solute + Mass of solvent = 3.658 g + 25.2 g = 28.858 g

Percentage by mass = (Mass of solute/Mass of solution) x 100



∴  Percentage by mass of urea = (3.658/28.858) x 100 = 12.68%

Molecular mass of methyl alcohol (CH3OH) =  12 g x 1 + 1 g x 4 + 16g  x 1 = 12 + 4 + 16 = 32 g mol-1

Number of moles of solute (methyl alcohol) = Given mass/ molecular mass

Number of moles of solute (methyl alcohol) = nB = 3.658 g/ 32 g = 0.1143 mol

Molecular mass of water (H2O) = nA = 1 g x 2 + 16g  x 1 = 2 + 16 = 18 g mol-1



Number of moles of solvent (water) = Given mass/ molecular mass

Number of moles of solvent (water) = nB = 25.2 g/ 18 g = 1.4 mol

Total number of moles = n+ n= 0.1143 + 1.4 = 1.5143 mol

Mole fraction of solute (methyl alcohol) = xB = nB/(n+ nB) = 0.1143/1.5143 = 0.0755

Mole fraction of solvent (water) = xA = nA/(n+ nB) = 1.2/1.5143 = 0.9245



Ans:  The percentage by mass of methyl alcohol is 12.68%

Mole fraction of methyl alcohol is 0.0755 and that of water is 0.9245

Example – 03:

  • Find the mole fraction of HCl in a solution of HCl containing 24.8 % of HCl by mass. Given H = 1, Cl = 35.5
  • Given: Percentage by mass = 24.8%
  • To Find: Mole fraction of HCl =?
  • Solution:

Percentage by mass of HCl = 24.8%

Let us consider 100 g of HCl solution

Mass of solute (HCl) = 24.8 g



Mass of solvent (water) = 100 – 24.8 = 75.2 g

The molecular mass of HCl = 35.5 g x 1 + 1 g x 1 = 36.5 g

Number of moles of solute (HCl) = Given mass/ molecular mass

Number of moles of solute (HCl) = nB = 24.8 g/ 36.5 g = 0.6795 mol



The molecular mass of water = 1 g x 2 + 16 g x 1 = 18 g

Number of moles of solvent (water) = Given mass/ molecular mass

Number of moles of solvent (water) = nA = 75.2 g/ 18 g = 4.178 mol

Total number of moles = n+ n= 4.178 + 0.6795 = 4.8575 mol

Mole fraction of solute (HCl) = xB = nB/(n+ nB) = 0.6795/4.8575 = 0.1399

Mole fraction of HCl = 0.1399



Example – 04:

  • Calculate the mol fraction of ethylene glycol (C2H6O2) in a solution containing 20% of (C2H6O2) by mass in aqueous solution.
  • Given: 20% of ethylene glycol (C2H6O2)
  • To Find: Mole fraction of ethylene glycol (C2H6O2) =?
  • Solution:

Molecular mass of ethylene glycol (C2H6O2) = 12 g x 2 + 1 g x 6 + 16 g x 2 = 62 g mol-1

Molecular mass of water (H2O) = 1 g x 2 + 16 g x 1 = 18 g mol-1

Let us consider 100 g of ethylene glycol (C2H6O2) solution

Mass of solute (ethylene glycol) = 20 g



Mass of solvent (water) = 100 – 20 = 80 g

Number of moles of solute (ethylene glycol) = nB = 20 g/ 62 g = 0.3226 mol

Number of moles of solvent (water) = nB = 80 g/ 18 g = 4.4444 mol

Total number of moles = n+ n= 4.444 + 0.3226 = 4.767 mol

Mole fraction of solute (ethylene glycol) = xB = nB/(n+ nB) = 0.3226/4.767 = 0.0677

Example – 05:

  • Calculate the mol fraction of benzene in a solution containing 30% by mass in carbon tetrachloride..
  • Given: 30% of benzene in carbon tetrachloride.
  • To Find: Mole fraction of benzene =?
  • Solution:

Molecular mass of benzene (C6H6) = 12 g x 6 + 1 g x 6  = 78 g mol-1



Molecular mass of carbon tetrachlorider (CCl4) = 12 g x 1 + 35.5 g x 1 = 154 g mol-1

Let us consider 100 g of the solution  (C6H+ CCl4)

Mass of solute (ethylene glycol) = 30 g

Mass of solvent (water) = 100 – 30 = 70 g

Number of moles of solute (benzene) = nB = 30 g/ 78 g = 0.3846 mol

Number of moles of solvent (carbon tetrachloride) = nB = 70 g/ 154 g = 0.4545 mol

Total number of moles = n+ n= 0.4545 + 0.3846 = 0.8391 mol

Mole fraction of solute (benzene) = xB = nB/(n+ nB) = 0.3846/0.8391 = 0.4583

Mole fraction of solute (carbon tetrachlopride) = xA =1 – 0.4583 = 0.5417

Example – 06:

  • A solution contains 25% water, 25% ethyl alcohol and 50% acetic acid by mass calculate the mole fraction of each component.
  • Given: 25% water, 25% ethyl alcohol and 50% acetic acid by mass
  • To Find: mole fraction of each constituent =?
  • Solution:

Let us consider 100 g of solution

Mass of water = 25 g, Mass of ethyl alcohol = 25 g and mass of acetic acid = 50 g

Molecular mass of water (H2O) = 1 g x 2 + 16 g x 1 = 18 g mol-1

Molecular mass of ethyl alcohol (C2H5OH) =  12 g x 2 + 1 g x 6 + 16g  x 1 = 46 g mol-1

Molecular mass of acetic acid (CH3COOH) =  12 g x 2 + 1 g x 4 + 16g  x 2 = 60 g mol-1

Number of moles of water = nA = 25 g/ 18 g = 1.3889 mol

Number of moles of ethyl alcohol = nB = 25 g/ 46 g = 0.5435 mol

Number of moles of acetic acid = nC = 50 g/ 60 g = 0.8333 mol

Total number of moles = nA + nB + nC = 1.3889 + 0.5435 + 0.8333 = 2.7657

Mole fraction of water = xA = nA/(n+n+ nC) = 1.3889/2.7657 = 0.5022

Mole fraction of ethyl alcohol = xB = nB/(n+n+ nC) = 0.5435/2.7657 = 0.1965

Mole fraction of acetic acid = xC = nC/(n+n+ nC) = 0.8333/2.7657 = 0.3013

Science > Chemistry > Solutions and Colligative PropertiesYou are Here
Physics Chemistry  Biology  Mathematics

Leave a Comment

Your email address will not be published. Required fields are marked *