PROBLEM SETS: %BY MASS, % BY VOLUME, MOLARITY, MOLALITY (show your solution)
1. What is the molarity of a solution in which 0.850 grams of ammonium nitrate are dissolved in 345 mL of solution?
2. Calculate the molality of a solution of 13.5g of KF dissolved in 250. g of water.
3. Calculate the molality of a solution containing 16.5 g of dissolved naphthalene (C10H8) in 54.3 g benzene (C6H6).
4. What is the mass percent of each component in the mixture formed by adding 12 g of calcium sulfate, 18 g of sodium nitrate, and 25 g of potassium chloride to 500 g of water?
5. A solution is made by dissolving 125 g of sodium chloride in 1.5 kg of water. What is the percent by mass?
6. What is the percent by volume of a solution formed by added 15 L of acetone to 28 L of water?

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Pstnonsonjokugo Pstnonsonjokugo · Answer 1

The concentration of a substance can be expressed using molarity, molality or percent.

The term concentration has to do with the amount of substance present in solution. Now let us solve the problems individually;

a) Number of moles = 0.850 grams/80 g/mol = 0.011 moles

molarity = 0.011 moles/345 * 10^-3 L = 0.032 M

b) Number of moles = 13.5g /58 g/mol = 0.23 moles

molality = 0.23 moles/250 * 10^-3 Kg = 0.92 m

c) Number of moles = 16.5 g/128 g/mol = 0.13 moles

molality = 0.13 moles/54.3 * 10^-3 Kg =2.39 m

d) Total mass present = 12 g + 18 g + 25 g + 500 g = 555 g

mass percent of calcium sulfate = 12 g/555 g * 100/1 = 2.2 %

mass percent of sodium nitrate = 18 g/ 555 g * 100/1 = 3.2 %

mass percent of potassium chloride = 25 g / 555 g * 100/1 = 4.5%

mass percent of water = 500 g / 555 g * 100/1 =90.1%

e) Total mass present = 125 g + 1500g = 1625 g

Mass percent of NaCl = 125 g/1625 g * 100/1 = 7.7%

f) Total volume of solution = 15 L + 28 L = 43 L

percent by volume of acetone = 15 L/43 L * 100/1 = 34.9%

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