Science > Chemistry > Solutions and Their Colligative Properties > Numerical Problems on Mole Fraction
In this article, we shall study to solve problems to calculate mole fraction of solute and solvent.
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 = nA + nB = 0.5 + 3 = 3.5 mol
Mole fraction of solute (ethyl alcohol) = xB = nB/(nA + nB) = 0.5/3.5 = 0.1429
Mole fraction of solvent (water) = xA = nA/(nA + nB) = 3/3.5 = 0.8571
Ans: Mole fraction of solute (ethyl alcohol) = 0.1429 and mole fraction of solvent (water) = 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 = nA + nB = 0.1143 + 1.4 = 1.5143 mol
Mole fraction of solute (methyl alcohol) = xB = nB/(nA + nB) = 0.1143/1.5143 = 0.0755
Mole fraction of solvent (water) = xA = nA/(nA + nB) = 1.2/1.5143 = 0.9245
Ans: The percentage by mass of methyl alcohol is 12.68% and 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 = nA + nB = 4.178 + 0.6795 = 4.8575 mol
Mole fraction of solute (HCl) = xB = nB/(nA + nB) = 0.6795/4.8575 = 0.1399
Ans: Mole fraction of HCl = 0.1399
Example – 04:
Calculate the mole 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 = nA + nB = 4.444 + 0.3226 = 4.767 mol
Mole fraction of solute (ethylene glycol) = xB = nB/(nA + nB) = 0.3226/4.767 = 0.0677
Ans: Mole fraction of ethylene glycol = 0.0677
Example – 05:
Calculate the mole 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 tetrachloride (CCl4) = 12 g x 1 + 35.5 g x 1 = 154 g mol-1
Let us consider 100 g of the solution (C6H6 + 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 = nA + nB = 0.4545 + 0.3846 = 0.8391 mol
Mole fraction of solute (benzene) = xB = nB/(nA + nB) = 0.3846/0.8391 = 0.4583
Ans: Mole fraction of benzene = 0.4583
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/(nA +nB + nC) = 1.3889/2.7657 = 0.5022
Mole fraction of ethyl alcohol = xB = nB/(nA +nB + nC) = 0.5435/2.7657 = 0.1965
Mole fraction of acetic acid = xC = nC/(nA +nB + nC) = 0.8333/2.7657 = 0.3013
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