Molarity and Dilutions practice PDF

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molarity and dilutions...

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Molarity Calculations

Name: Alexander Krupinski

Calculate the molarities of the following solutions: 1)

2.3 moles of sodium chloride (NaCl) in 0.45 liters of solution.

moles of NaCl = 2.3 mol L = 0.45 L M = 5.1 mol/L

2)

1.2 moles of calcium carbonate (CaCO3) in 1.22 liters of solution.

moles of CaCO3 = 1.2 mol L = 1.22 L M = 0.98 mol/L

3)

0.09 moles of sodium sulfate (Na2SO4) in 12 mL of solution.

moles of Na2SO4 = 0.009 mol L = 0.012 L M = 0.75 mol/L

4)

0.75 moles of lithium fluoride (LiF) in 65 mL of solution.

moles of LiF = 0.75 mol L = 0.065 M = 11.54 mol/L

5)

0.8 moles of magnesium acetate (Mg(C2H3O2) 2) in 5 liters of solution.

moles of Mg(C2H3O2)2 = 0.8 mol L=5L M = 0.16 mol/L

6)

120 grams of calcium nitrite (Ca(NO2)2) in 240 mL of solution.

moles of Ca(NO2)2 = 0.73 mol L = 0.24

M = 3.042 mol/L

7)

98 grams of sodium hydroxide (NaOH) in 2.2 liters of solution.

moles of NaOH = 2.45 mol L = 2.2 L M = 1.1 mol/L

8)

1.2 grams of hydrochloric acid (HCl) in 25 mL of solution.

moles of HCl = 0.033 mol L = 0.025 L M = 1.32 mol/L

9)

45 grams of ammonia (NH3) in 0.75 L of solution.

moles of NH3 = 2.64 mol L = 0.75 L M = 3.52 mol/L

Dilutions M1V1 = M2V2 1. If I dilute 250 mL of a 0.10 M CaCl2 solution to a volume of 750 mL, what will the concentration of this solution be? M1 = 0.10 mol/L CaCl2 V1 = 0.25 L V2 = 0.75 L 0.10 mol/L * 0.25 L = M2 * 0.75 L M2 = (0.10 mol/L * 0.25 L) / 0.75 L = 0.03 Answer: 0.03 mol/L

2. If I need to make 225 mL of a 0.5 M LiCl solution, how much of the original 2 M LiCl solution do I use? M1 = 2 mol/L LiCl V1 = 0.225 L M2 = 0.5 mol/L LiCl 2 mol/L * 0.255 L = 0.5 mol/L * V2 V2 = (2 mol/L * 0.225 L) / 0.5 mol/L = 0.9 L

Answer: 0.9 L

3. If I take 35 mL of a 3.0 M Na2SO4 solution and dilute it to a 750 mL solution, what is the new concentration? M1 = 3.0 mol/L Na2SO4 V1 = 0.035 L V2 = 0.75 L 3.0 mol/L * 0.035 L = M2 * 0.75 L M2 = (3.0 mol/L * 0.035 L) / 0.75 L = 0.14 mol/L Answer: 0.14 mol/L

4. If I want to dilute 500 mL of a 2 M PbNO3 solution to a 1.5 M PbNO3 solution, what will be the volume? M1 = 2 mol/L PbNO3 V1 = 0.5 L M2 = 1.5 mol/L PbNO3 2 mol/L * 0.5 L = 1.5 mol/L * V2 V2 = (2 mol/L * 0.5 L) / 1.5 mol/L = 0.6 L Answer: 0.6 L

5. If I need to make 6.5 L of a 1.0 M CuSO4 solution, how much of the original 4.5 M CuSO4 solution will I need? M1 = 4.5 mol/L CuSO4 V1 = 6.5 L M2 = 1.0 mol/L CuSO4 4.5 mol/L * 6.5 L = 1.0 mol/L * V2

V2 = (4.5 mol/L * 6.5 L) / 1.0 mol/L = 29.25 L Answer: 29.25 L

6. If I want 250 mL of a 1.0 M KCl solution, how much of the original 5.0 M KCl solution should I start with? M1 = 5.0 mol/L KCl V1 = 0.25 L M2 = 1.0 mol/L KCl 5.0 mol/L * 0.25 L = 1.0 mol/L * V2 V2 = (5.0 mol/L * 0.25 L) / 1.0 mol/L = 1.25 L Answer: 1.25 L

7. If I want to dilute 30 mL of a 1.5 M FeSO4 solution to a 0.3 M FeSO4 solution, how much of the solvent will I need to add? M1 = 1.5 mol/L FeSO4 V1 = 0.03 L M2 = 0.3 mol/L FeSO4 1.5 mol/L * 0.03 L = 0.3 mol/L * V2 V2 = (1.5 mol/L * 0.03 L) / 0.3 mol/L = 0.15 L Answer: 0.15 L

8. If I have 340 mL of a 0.5 M NaBr solution, what will the concentration be if I add 560 mL more water to it? M1 = 0.5 mol/L NaBr V1 = 0.34 L V2 = 0.9 L 0.5 mol/L * 0.34 L = M2 * 0.9 L M2 = (0.5 mol/L * 0.34 L) / 0.9 L = 0.18 mol/L Answer: 0.18 mol/L

9. If I want to dilute 47 mL of 18 M H2SO4 to 1 M H2SO4, how much water will I need to add? M1 = 18 mol/L H2SO4 V1 = 0.047 L M2 = 1 mol/L H2SO4 18 mol/L * 0.047 L = 1 mol/L * V2 V2 = (18 mol/L * 0.047 L) / 1 mol/L = 0.846 L Answer: 0.846 L

10. If I leave 750 mL of a 0.75 M NaCl solution out and 150 mL of the solvent evaporates, what is the new concentration of the NaCl solution? M1 = 0.75 mol/L NaCl V1 = 0.75 L V2 = 0.6 L 0.75 mol/L * 0.75 L = M2 * 0.6 L M2 = (0.75 mol/L * 0.75 L) / 0.6 L = 0.9375 mol/L Answer: 0.9375 mol/L...