Exercises
Exercise 1. What is the concentration, M, of an aqueous solution prepared by adding enough solvent to 4.75 g of CaCl2 to make 250.00 mL of solution.
We need to convert 4.75 g of CaCl2 to moles–0.0428 moles. We then convert the 250.00 mL of solution to liters–0.25000 L.
Exercise 2. How many moles of glucose are present in 155 mL of 0.85 M glucose (C6H12O6)?
155 mL = 0.155 L
\(\displaystyle moles\;C_6H_{12}O_6\;=\;0.155\;L\;\times\;0.85\;\frac{mol}{L}\;=\;\mathbf{0.132\;mol\;C_6H_{12}O_6}\)
Exercise 3. How many grams of boric acid (H3BO3) are present in 245 mL of 0.450 M H3BO3?
Mm H3BO3 = 61.83 g/mol
\(\displaystyle moles\;H_3BO_3\;=\;0.245\;L\;times\;0.450\;\frac{mol}{L}\;=\;0.11\underline{0}25\;mol\;H_3BO_3\)Convert moles to g
\(\displaystyle 0.11\underline{0}25\;mol\;\times\;\frac{61.83\;g}{1\;mol}\;=\;\mathbf{6.82\;g\;H_3BO_3}\)Exercise 4. How many mL of 0.145 M NaCl solution contain 2.35 g of NaCl?
First convert g of NaCl to moles. Mm NaCl = 58.44 g/mol
\(\displaystyle 2.35\;g\;NaCl\;\times\;\frac{1\;mol\;NaCl}{58.44\;g\;NaCl}\;=\;{0.040\underline{2}12\;mol\;NaCl}\)
\(\displaystyle V\;=\;\frac{0.040\underline{2}12\;mol}{0.145\;M}\;=\;\mathbf{0.277\;L}\)
0.277 L = 277 mL
Exercise 5. What volume of solution, in mL, would result from the dilution of 25.00 mL of 3.50 M NaOH to 0.100 M NaOH?
Here we use M1V1 = M2V2 and solve for M2
M1 = 3.50 M NaOH, V1 = 25.00 mL, M2 = ?, and V2 = 50.00 mL
\(\displaystyle M_2\;=\;\frac{3.50\;M\;\times\;25.00\;mL}{50.00\;mL}\;=\;\mathbf{1.75\;M}\)
Exercise 6. Calculate the final volume if 5.00 mL of 1.65 M CaCl2 is diluted to a final concentration of 0.105 M.
Here we use M1V1 = M2V2 and solve for V2
M1 = 1.65 M, V1 = 5.00 mL, M2 = 0.105 M, and V2 = ?
\(\displaystyle V_2\;=\;\frac{1.65\;M\;\times\;5.00\;mL}{0.105\;M}\;=\;\mathbf{78.6\;mL}\)