# Problems on Percentage by Mass and Volume

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### Problems on Percentage by Mass

#### Example – 01:

• 6 g of urea was dissolved in 500 g of water. calculate the percentage by mass of urea in the solution.
• Given: Mass of solute (urea) = 6 g, Mass of solvent (water) = 500 g
• To Find: Percentage by mass = ?
• Solution:

Mass of solution = Mass of solute + Mass of solvent = 6 g + 500 g = 506 g

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

∴  Percentage by mass of urea = (6/506) x 100 = 1.186%

#### Example – 02:

• 34.2 g of glucose is dissolved in 400 g of water. Calculate the percentage by mass of glucose solution.
• Given: Mass of solute (glucose) = 34.2 g, Mass of solvent (water) = 400 g
• To Find: Percentage by mass =?
• Solution:

Mass of solution = Mass of solute + Mass of solvent = 34.2 g + 400 g = 434.2 g

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

Percentage by mass of glucose = (34.2/434.2) x 100 = 7.877%

#### Example – 03:

• A solution is prepared by dissolving 15 g of cane sugar in 60 g of water. Calculate the mass per cent of each component of the solution.
• Given: Mass of solute (cane sugar) = 15 g, Mass of solvent (water) = 60 g
• To Find: Mass per cent of cane sugar and water =?
• Solution:

Mass of solution = mass of solute + mass of solvent = 15 g + 60 g = 75 g

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

Percentage by mass of solute (cane sugar) = (15 g/75 g) x 100 = 20%

Percentage by mass of solvent (water) = 100 – 20 = 80%

#### Example – 04:

• Calculate the mass percentage of benzene and carbon tetrachloride if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
• Given: Mass of solute (benzene) = 22 g, Mass of solvent (carbon tetrachloride) = 122 g.
• To Find: Mass percentage of benzene and carbon tetrachloride.
• Solution:

Mass of solution = mass of solute + mass of solvent

Mass of solution = 22 g + 122 g = 144 g

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

Percentage of benzene by mass = (22 g/144 g) x 100 = 15.28%

Percentage of carbon tetrachloride by mass = 100 – 15.28 = 84.72%

#### Example – 05:

• A solution is prepared by dissolving a certain amount of solute in 500 g of water. The percentage by mass of a solute in a solution is 2.38. Calculate the mass of solute
• Given:  Mass of solvent = 400 g, percentage by mass = 2.38
• To Find: Mass of solute =?
• Solution:

Let the mass of solute = x g

Mass of solution = Mass of solute + Mass of solvent = x g + 500 g = (x + 500) g

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

2.38 = (x g/(x + 500) g) x 100

2.38 (x + 500) = 100x

2.38x + 1190 = 100x

1190 = 97.62 x

x = 1190/97.62 = 12.19 g

The mass of solute is 12.19 g

#### Example – 06:

• Caculate the masses of cane sugar and water required to prepare 250 g of 25% cane sugar solution.
• Given: 250 g of 25% cane sugar solution
• To Find: Masses of cane sugar and water =?
• Solution:

Let the mass of cane sugar = x g

Mass of solution = 250 g

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

25 = (x g/250 g) x 100

25 x 250 g = 100x

6250 g = 100x

x = 6250 g/100 = 62.5 g

Mass of cane sugar = 62.5 g

Mass of water = 250 g – 62.5 g = 187.5 g

#### Example – 07:

• 15 g of methyl alcohol is present in 100 mL of solution. If the density of solution is 0.96 g mL-1. calculate the mass percentage of methyl alcohol solution.
• Given: Mass of solute (methyl alcohol) = 15 g, Volume of solution = V = 100 mL, Density of solution = d = 0.96 g mL-1.
• To Find: mass percentage of methyl alcohol =?
• Solution:

Mass of solution = volume x density = 100 mL x 0.96 g mL-1 = 96 g

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

Percentage of benzene by mass = (15 g/96 g) x 100 = 15.625%

#### Example – 08:

• The density of solution of salt X is 1.15 g mL-1. 20 mL of the solution when completely evaporated gave a residue of 4.6 g of the salt. Calculate the mass percentage of solute in the solution.
• Given: Volume of solution = V = 20 mL, density of solution = d = 1.15 g mL-1, Mass of solute = 4.6 g
• To Find: mass percentage of solute in the solution = ?
• Solution:

Mass of solution = volume x density = 20 mL x 1.15 g mL-1 = 23 g

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

Percentage of solute by mass = (4.6 g/23 g) x 100 = 20%

#### Example – 09:

• 40% by mass of urea is obtained when 190 g of urea is dissolved in 400 mL of water. Calculate the density of solution.
• Given: % by mass of urea solution = 40%, mass of solvent (water)  = 400 mL.
• To Find: Density of solution =?
• Solution:

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

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

Mass of solution = (190 g/40) x 100 = 475 g

The volume of solvent (water) = 400 mL = Volume of solution

Density of solution = mass of solution /volume of solution

Density of solution = (475 g)/(400 mL) = 1.19 g mL-1

#### Example – 10:

• Calculate percentage composition in terms of mass of a solution obtained by mixing 300 g of 25% solution of NH4NO3 with 400 g of a 40% solution of solute X.
• Given: 300 g of 25% solution of NH4NOmixed with 400 g of a 40% solution of solute X
• To Find: percentage composition in terms of mass =?
• Solution:

Consider 300 g of 25% solution of NH4NO3

Mass of solute in this solution = 25% of 300 g = (25/100) x 300 g = 75 g

Consider 400 g of a 40% solution of solute X

Mass of solute in this solution = 40% of 400 g = (40/100) x 400 g = 160 g

Now let us consider the solution obtained by mixing

Total mass of solute = WB = 75 g + 160 g = 235 g

Total mass of solution = WA = 300 g + 400 g = 700 g

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

Percentage of solute by mass = (235 g/700 g) x 100 = 33.57%

Percentage of solvent by mass = 100 – 33.57 = 66.43%

#### Example – 11:

• Calculate percentage composition in terms of mass of a solution obtained by mixing 100 g of 30% solution of NaOH with 150 g of a 40% solution of NaOH.
• Given: 100 g of 30% solution of NaOH  mixed with 150 g of a 40% solution of NaOH
• To Find: percentage composition in terms of mass =?
• Solution:

Consider 100 g of 30% solution of NaOH

Mass of solute in this solution = 30% of 100 g = (30/100) x 100 g = 30 g

Consider 150 g of a 40% solution of NaOH

Mass of solute in this solution = 40% of 150 g = (40/100) x 150 g = 60 g

Now let us consider the solution obtained by mixing

Total mass of solute = WB = 30 g + 60 g = 90 g

Total mass of solution = WA = 100 g + 150 g = 250 g

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

Percentage of solute by mass = (90 g/250 g) x 100 = 36%

Percentage of solvent by mass = 100 – 36 = 64%

### Problems on Percentage by Volume

#### Example – 12:

• 12.8 cm3 of benzene is dissolved in 16.8 cm3 of xylene. Calculate percentage by volume of benzene.
• Given: Volume of solute = 12.8 cm3, Volume of solvent =  16.8 cm3
• To Find: Percentage by volume =?
• Solution:

Volume of solution = Volume of solute + Volume of solvent

Volume of solution = 12.8 cm3+ 16.8 cm3  =  29.6 cm3

Percentage by volume = (Volume of solute/Volume of solution) x 100

Percentage of benzene by volume = (12.8 cm3/29.6 cm3) x 100 = 43.24 %

#### Example – 13:

• 58 cm3 of ethyl alcohol was dissolved in 400 cm3 of water to form 454 cm3 of a  solution of ethyl alcohol. Calculate percentage by volume of ethyl alcohol in water. (12.78 % by volume)
• Given: Volume of solute = 58 cm3, Volume of solution =  454 cm3
• To Find: Percentage by volume =?
• Solution:

Percentage by volume = (Volume of solute/Volume of solution) x 100

Percentage of ethyl alcohol by volume = (58 cm3/454 cm3) x 100 = 12.78%

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