Solutions: Mass Percent - Video Tutorials & Practice Problems
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Mass Percent represents grams of solute per grams of solution, multiplied by 100.
Mass Percent of Solutions
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Solutions: Mass Percent Concept 1
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For solutions, mass percent represents the grams of solute per grams of solution multiplied by 100. So when we're looking at the mass percent of solution, just remember that it equals grams of solute divided by grams of solution multiplied by 100. So keep this formula in mind anytime we're given a question that deals with solutions and mass percent.
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Solutions: Mass Percent Example 1
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Calculate the mass percent of a solution prepared by dissolving 34.1 grams of potassium chloride in 150 grams of water. Alright. So we're going to say here that our mass percent equals the grams of solute. So remember the solute is the one that's smaller in amount. In this case, it'd be the 34.1 grams of potassium chloride divided by grams of solution. Remember, your solution is your solute and your solvent added together. So kcl is our solute plus water, which is our solvent, and that'd be times 100. When you plug this in, you'll get back as your mass percent, 18.5%. So this will represent the mass percent of our given solution.
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Problem
Problem
Calculate the amount of water (in kilograms) that must be added to 12.0 g of urea, (NH2)2CO, in the preparation of a 18.3 percent by mass solution. The molar mass of urea, (NH2)2CO, is 60.055 g/mol.
A
0.644 kg
B
0.0535 kg
C
6.44 kg
D
0.0755 kg
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Solutions: Mass Percent Concept 2
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51s
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Now it's important to remember that when converting to other unit terms such as molarity, the mass percent value itself can represent a conversion factor. So here, for example, when we have 2.50% sodium hydroxide solution, it can be represented as 2.50 grams of sodium hydroxide within a 100 grams of solution. So this can be written as 2.50 grams NaOH over 100 grams of solution. Now since it's a conversion factor, we can also do the reciprocal where we have 100 grams of solution on top and 2.50 grams of NaOH on the bottom. So keep this in mind, we can basically expand what the mass percent is whenever a value is given to us. That percentage is that number over 100 grams of solution.
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Solutions: Mass Percent Example 2
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Determine the molarity of a sulfuric acid solution that is 5% sulfuric acid with a density of 0.9918 grams per milliliter. Alright. So remember, molarity itself is capital m. So we need to isolate the units of moles over liters in order to isolate molarity. Alright. So remember that when we are given a mass percent, it can be used as a conversion factor as well. So 5% sulfuric acid here really means 5 grams of sulfuric acid per 100 grams of solution. And here, we have the density of the solution, which is 0.9918 grams of solution per 1 milliliter of solution. Alright. So how do we use this information to help us get to our molarity? Alright. So we're gonna start out with our mass percent, which is 5.0 grams of sulfuric acid per 100 grams of solution. Alright. And we're gonna say here, we wanna get rid of the grams of sulfuric acid and isolate moles of sulfuric acid. So for every 1 mole of sulfuric acid, the molar mass of it is 98.086 grams. Grams here cancel out. Now I have moles of sulfuric acid over grams of solution. Next, I need to get rid of grams of solution because remember, molarity has liters on the bottom. We do this by using our density. So here, our density is 0.9918 grams of solution per 1 milliliter of solution. So here grams of solution cancel out, and now I have milliliters of solution. I'm almost there. All I gotta do now is just change these milliliters into liters. So remember that 1 milli is equal to 10 to the negative 3 liters. And like that we've isolated moles over liters. So we've just isolated our molarity. So when we plug that in, we're gonna get 0.5056 molar. Now since this has 2 sig figs, we'll just bring this down to 2 sig figs as well. So this is 0.51 molar. So that'll be our molarity for this given solution of sulfuric acid.
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Problem
Problem
A solution was prepared by dissolving 51.0 g of KBr in 310 mL of water. Calculate the mass percent of KBr in the solution.
A
16.4%
B
14.1%
C
19.7%
D
0.141%
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Problem
Problem
An aqueous LiNO2 solution is made using 90.3 g LiNO2 and diluting it to a total volume of 1.72 L. If the density of the solution is 1.20 g/mL, what is the mass percent of the solution?
A
4.37%
B
5.25%
C
4.19%
D
4.99%
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Problem
Problem
Determine the percent sulfuric acid by mass of a 1.37 m aqueous solution of H2SO4.