Molarity practice 1 PDF

Preview

Summary

molarity practice...

Description

Molarity Practice 1) How many grams of potassium chloride are needed to make 200 mL of a 2.5 M solution? M = 2.5 mol/L L = 200 mL = 0.2 L Molar mass of potassium chloride = 74.6 g mol = 0.2 L * 2.5 mol/L * 74.6 g = 37.3 g KCl Answer: 37.3 g KCl

2) How many liters of 5 M solution can be made using 120 grams of sodium bromide? M = 5 mol/L g of NaBr = 120 Molar mass of NaBr = 103 g L = (120 g / 103 g) / 5 mol/L = 0.223 L Answer: 0.223 L

3) What is the concentration of an aqueous solution with a volume of 550 mL that contains 150 grams of iron (II) chloride? L = 0.55 L g of iron (II) chloride = 150 g Molar mass of iron (II) chloride = 127 g M = (150 g / 127 g) / 0.55 L = 2.145 mol/L Answer: 2.145 mol/L

4) How many grams of ammonium phosphate are needed to make a 300ml solution at a concentration of 4 M? L = 0.3 L

M = 4 mol/L Molar mass of ammonium phosphate = 149.1 g of ammonium phosphate = (0.3 L * 4 mol/L) * 149.1 = 178.92 g Answer: 178.92 g of ammonium phosphate

5) What is the concentration of a solution with a volume of 3.5 liters containing 560 grams of calcium phosphate? L = 3.5 L g of CaP = 560 g Molar mass of CaP = 71.1 g M = (560 g / 71.1 g) / 3.5 L = 2.25 mol/L Answer: 2.25 mol/L

6) How many grams of copper (II) fluoride are needed to make 6.7 liters of a 1.2 M solution? M = 1.2 mol/L L = 6.7 L Molar mass of copper (II) fluoride = 102 g g of copper (II) fluoride = (6.7 L * 1.2 mol/L) * 102 g = 820.08 g Answer: 820.08 of copper (II) fluoride

7) How many liters of a 0.65 M solution can be made with 15.5 grams of lithium fluoride? M = 0.65 mol/L g of LiF = 15.5 g Molar mass of LiF = 26 g L = (15.5 g / 26 g) / 0.65 mol/L = 0.917 L Answer: 0.917 L

8) What is the concentration of a solution with a volume of 360 mL that contains 35.4 grams of aluminum acetate? L = 0.36 L g of aluminum acetate = 35.4 g Molar mass of aluminum acetate = 204 g M = (35.4 g / 204 g) / 0.36 L = 0.48 mol/L Answer: 0.48 mol/L

9) How many liters of a 0.55 M solution can be made with 85 grams of lead (II) oxide? M = 0.55 mol/L Molar mass of PbO = 223.2 g g of PbO = 85 g L = (85 g / 223.2 g) / 0.55 mol/L = 0.692 L Answer: 0.692 L

10) How many grams of manganese (IV) oxide are needed to make 3.6 liters of a 5.1 M solution? M = 5.1 mol/L L = 3.6 L Molar mass of MnO2 = 87 g g of MnO2 = (5.1 mol/L * 3.6 L) * 87 g = 1597.32 g Answer: 1597.32 g of MnO2

11) What is the concentration of a solution with a volume of 20 mL that contains 5 grams of iron (III) hydroxide? L = 0.02 L g of iron (III) hydroxide = 5 g

Molar mass of iron (III) hydroxide = 107 g M = (5 g / 107 g) / 0.02 L = 2.34 mol/L Answer: 2.34 mol/L

12) How many liters of a 3.4 M isopropanol solution can be made with 78 grams of isopropanol (C3H8O)? M = 3.4 mol/L g of C3H8O = 78 g Molar mass of C3H8O = 60.1 g L = (78 g / 60.1 g) / 3.4 mol/L = 0.38 L Answer: 0.38 L

13) What is the concentration of a solution with a volume 5.3 mL that contains 15 grams of ammonium sulfite? L = 0.0053 Molar mass of ammonium sulfite = 116 g g of ammonium sulfite = 15 g M = (15 g / 116 g) / 0.0053 L = 24.4 mol/L Answer: 24.4 mol/L...