In this article, us shall find out to calculate percentages by mass and percentage by volume the a solution.
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Example – 01:
6 g that urea was liquified in 500 g of water. Calculation the percent by fixed of urea in the solution.
Given: mass of solute (urea) = 6 g, mass of solvent (water) = 500g
To Find: Percent by mass =?
Solution:
Mass of solution = mass of solute + fixed of solvent = 6 g +500 g = 506 g
Percentage by massive of urea = (Mass the solute/Mass of solution) x 100
= (6/506) x 100 = 1.186%
Example – 02:
34.2 g that glucose is liquified in 400 g the water. Calculation the percentage by massive of glucose solution.
Given: massive of solute (glucose) = 34.2 g, fixed of solvent (water)= 400 g
ToFind: percent by massive =?
Solution:
Mass of systems = fixed of solute + fixed of solvent = 34.2 g+ 400 g = 434.2 g
Percentage by mass = (Mass of solute/Mass the solution) x 100
= (34.2/434.2) x 100 = 7.877%
Example – 03:
A systems is all set by dissolve 15 g that cane sugar in 60 g of water. Calculate the mass percent of each component that the solution.
Given: massive of solute (cane sugar) = 15 g, massive of solvent (water)= 60 g
To Find: fixed percent of cane sugar and water =?
Solution:
Mass of systems = mass of solute + mass of solvent = 15 g +60 g = 75 g
Percentage by massive of solute c(cane sugar) = (Mass ofsolute/Mass that solution) x 100
Mass percent the solute (cane sugar) = (15 g/75 g) x 100 = 20%
Mass percent of solvent (water) = 100 – 20 = 80%
Example – 04:
Calculate the mass percentage of benzene and also carbon tetrachloride if 22 g the benzene is dissolved in 122 g of carbon tetrachloride.
Given: massive of solute (benzene) = 22 g, massive of solvent (carbontetrachloride) = 122 g.
ToFind: Mass percent of benzene andcarbon tetrachloride.
Solution:
Mass of solution = mass of solute + massive of solvent
Mass of solution = 22 g + 122 g = 144 g
Percentage by fixed = (Mass the solute/Mass the solution) x 100
Percentage the benzene by mass = (22 g/144 g) x 100 = 15.28%
Percentage the carbon tetrachloride by mass = 100 – 15.28 =84.72%
Example – 05:
A solution is all set by dissolving a details amount that solute in 500 g that water. The percentage by massive of a solute in a solution is 2.38. Calculation the mass of solute
Given: massive of solvent = 400 g, portion by massive = 2.38
ToFind: massive of solute =?
Solution:
Let the mass of solute = x g
Mass of systems = fixed of solute + fixed of solvent = x g +500 g = (x + 500) g
Percentage by fixed = (Mass of solute/Mass that 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:
Calculate the masses of cane sugar and also water forced to prepare 250 g that 25% cane street solution.
Given:250 g that 25% cane sugar solution
ToFind: Masses the cane sugar and also water =?
Solution:
Let the massive of cane sugar = x g
Mass of equipment = 250 g
Percentage by mass = (Mass the 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 street = 62.5 g
Mass the water = 250 g – 62.5 g = 187.5 g
Example – 07:
15 g that methyl alcohol is existing in 100 mL the solution. If the density of solution is 0.96 g mL-1. Calculation the mass portion of methyl alcohol solution.
Given: mass of solute (methyl alcohol) = 15 g, Volume that solution= V = 100 mL, density of systems = d =0.96 g mL-1.
ToFind:mass percent of methyl alcohol =?
Solution:
Mass of solution = volume x density = 100 mL x 0.96 g mL-1=96 g
Mass portion = (Mass the solute/Mass that solution) x 100
Mass percent of benzene = (15 g/96 g) x 100 = 15.625%
Example – 08:
The thickness of the solution of salt X is 1.15 g mL-1. 20 mL that the systems when fully evaporated gave a residue that 4.6 g the the salt. Calculation the mass percentage of solute in the solution.
Given: Volume of equipment = V = 20 mL, thickness of equipment = d =1.15 g mL-1, fixed of solute = 4.6 g
ToFind:mass percentage of solute inthe systems =?
Solution:
Mass of systems = volume x thickness = 20 mL x 1.15 g mL-1=23 g
Percentage by mass = (Mass the solute/Mass the solution) x 100
Percentage of solute by fixed = (4.6 g/23 g) x 100 = 20%
Example – 09:
40% by massive of urea is obtained when 190 g of urea is liquified in 400 mL that water. Calculation the thickness of solution.
Given: % by fixed of urea solution = 40%, massive of solvent(water)= 400 mL.
ToFind: thickness of systems =?
Solution:
Percentage by fixed = (Mass of solute/Mass the solution) x 100
Mass of solution = (Mass that solute/Percentage by mass) x 100
Mass of solution = (190 g/40) x 100 = 475 g
The volume that solvent (water) = 400 mL = Volume the solution
Density of systems = fixed of equipment /volume that solution
Density of equipment = (475 g) / (400 mL) = 1.19 gmL-1
Example – 10:
Calculate percent ingredient in terms of mass the a solution obtained by mix 300 g the 25% solution of NH4NO3 v 400 g of a 40% solution of solute X.
Given:300 g of 25% solution of NH4NO3mixedwith400 g that a 40% solution of solute X
ToFind:percentage composition interms of massive =?
Solution:
Consider 300 g of 25% systems of NH4NO3
Mass of solute in this solution = 25% of 300 g = (25/100) x300 g = 75 g
Consider 400 g that a 40% systems of solute X
Mass of solute in this equipment = 40% of 400 g = (40/100) x400 g = 160 g
Now let us take into consideration the solution acquired by mixing
Total fixed of solute = WB = 75 g + 160 g = 235 g
Total fixed of solution = WA = 300 g + 400 g = 700g
Percentage by massive = (Mass that solute/Mass the solution) x 100
Percentage the solute by massive = (235 g/700 g) x 100 = 33.57%
Percentage the solvent by massive = 100 – 33.57 = 66.43%
Example – 11:
Calculatepercentage composition in terms of mass of a solution derived by mixing 100 gof 30% systems of NaOH v 150 g that a 40% solution of NaOH.
See more: Convert 49 Canadian To Us D, 49 Canadian Dollar To Us Dollar Exchange Rate
Given:100 g the 30% equipment of NaOHmixed with150 g ofa 40% solution of NaOH
ToFind:percentage composition interms of mass =?
Solution:
Consider 100 g of 30% equipment of NaOH
Mass of solute in this equipment = 30% the 100 g = (30/100) x100 g = 30 g
Consider 150 g that a 40% systems of NaOH
Mass of solute in this systems = 40% of 150 g = (40/100) x150 g = 60 g
Now allow us take into consideration the solution derived by mixing
Total mass of solute = WB = 30 g + 60 g = 90 g
Total fixed of systems = WA = 100 g + 150 g = 250g
Percentage by massive = (Mass that solute/Mass of solution) x 100
Percentage that solute by fixed = (90 g/250 g) x 100 = 36%
Percentage the solvent by mass = 100 – 36 = 64%
Problems on portion by Volume:
Example – 12:
12.8 cm3 of benzene is dissolved in16.8 cm3 of xylene. Calculate percent by volume of benzene.
Given: Volume that solute =12.8 cm3, Volume ofsolvent =16.8 cm3
ToFind: percent by volume =?
Solution:
Volume of systems = Volume that solute + Volume that solvent
Volume of systems = 12.8 cm3+ 16.8 cm3=29.6 cm3
Percentage through volume = (Volume of solute/Volume of solution)x 100
Percentage the benzene by volume = (12.8 cm3/29.6cm3) x 100 = 43.24 %
Example – 13:
58 cm3 of ethyl alcohol was liquified in 400 cm3 of water to type 454 cm3 the a solution of ethyl alcohol. Calculate portion by volume the ethyl alcohol in water. (12.78 % through volume)
Given: Volume that solute =58 cm3, Volume ofsolution =454 cm3
ToFind: portion by volume =?
Solution:
Percentage through volume = (Volume that solute/Volume of solution)x 100
Percentage the ethyl alcohol through volume = (58 cm3/454 cm3) x 100 = 12.78%
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