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 soluble programs of solution Multiplied by 100. So when we're looking at the mass percent of solutions, just remember that it equals grams of solute divided by grams of solution multiplied by 100. So keep this formula in mind any time 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% 34.1 g of potassium chloride in 150 g of water. All right, so we're going to say here that are mass percent equals the grams of solute. Remember, the solute is the one that's smaller amount in this case would be the 34.1 g of potassium chloride divided by grams of solution. Remember, your solution is your salute and your solvent added together. So K C. L is our salute. Mhm plus water, which is our solvent. And there will be times 100. When you plug this in, you'll get back as your mass% 18.5%. So this would 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, (NH_{2})_{2}CO, in the preparation of a 18.3 percent by mass solution. The molar mass of urea, (NH_{2})_{2}CO, 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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Now it's important to remember that when converting to other unit terms such as polarity, the mass percent value itself can represent a conversion factor. So here, for example, we have 2.50% sodium hydroxide solution. It can be represented as 2.50 g of sodium hydroxide within 100 g of solution, so this can be written as 2.50 g n a O h. Over 100 g of solution. Now, since it's a conversion factor, we can also do the reciprocal where we have 100 g of solution on top and 2.50 g of NH on the bottom. So keep this in mind. We can basically expand what the mass percent is whenever value is given to us. That percentage is that number over 100 g of solution.

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Solutions: Mass Percent Example 2

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determine the polarity of a sulfuric acid solution that is 5% sulfuric acid with a density of 50.9918 g per mole leader. Alright, so remember, Polarity itself is capital m. So we need to isolate the units of moles over leaders in order to isolate polarity. Alright. So remember that when we are given a mass%, 5 sulfuric acid here really means 5g of sulfuric acid Per 100 g of solution. And here we have the density of the solution, which is .9918 g of solution per one mL of solution. All right, so how do we use this information to help us get to our polarity? All right, so we're gonna start out with our mass%, 5.0 g of sulfuric acid Per 100 g of solution. What? All right. And we're gonna say here we want to get rid of the grams of sulfuric acid and isolate moles of sulfuric acid. So for every one mole of sulfuric acid, The molar mass of it is 98. g grams. You're cancel out now. I have Mosul sulfuric acid over grams of solution. Next, I need to get rid of grams of solution. Because remember, Polarity has leaders on the bottom. We do this by using our density. So here are density is .9918g of solution Per one million 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 leaders. So remember that one Milli is equal to 10 to the negative three leaders. And like that, we've isolated moles over liters. So we've just isolated are more clarity. So we plug that in, we're gonna get .5056 Mohler. Now, since this has two sig figs will just bring this down into six things as well. So this is 60.51 Moeller. So that'll be our polarity 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 LiNO_{2} solution is made using 90.3 g LiNO_{2} 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 H_{2}SO_{4}.

A

0.1342%

B

11.85%

C

13.42%

D

13.98%

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