Question

Complete the following table for aqueous solutions of copper(II) sulfate. $$ \begin{array}{cccc} \hline & \text { Mass of Solute } & \text { Volume of Solution } & \text { Molarity } \\ \hline \text { (a) } & 12.50 \mathrm{~g} & 478 \mathrm{~mL} & \\ \text { (b) } & \- & 283 \mathrm{~mL} & 0.299 \mathrm{M} \\ \text { (c) } & 4.163 \mathrm{~g} & \- & 0.8415 \mathrm{M} \\ \hline \end{array} $$

Short answer

Answer: The missing variables for each case are as follows: (a) Molarity = 0.1638 M (b) Mass of solute = 13.51 g (c) Volume of solution = 31.0 mL

Step-by-step solution

Calculate moles of solute for (a) and (c)

Use the mass of solute and molar mass of copper(II) sulfate (159.61 g/mol) to determine the moles of solute. For (a): \(\text{Moles of solute} = \frac{\text{Mass of solute}}{\text{Molar mass}} = \frac{12.50 \text{ g}}{159.61 \text{ g/mol}} = 0.0783 \text{ mol}\) For (c): \(\text{Moles of solute} = \frac{\text{Mass of solute}}{\text{Molar mass}} = \frac{4.163 \text{ g}}{159.61 \text{ g/mol}} = 0.0261 \text{ mol}\)

Calculate the molarity for (a) and volume of solution for (c)

Using the molarities and moles of solutes, we can find the missing variable in (a) and (c). For (a): Molarity = \(\frac{\text{Moles of solute}}{\text{Liters of solution}}\) \(\implies \text{Molarity} = \frac{0.0783 \text{ mol}}{478 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}}} = 0.1638 \text{ M}\) For (c): \(\text{Moles of solute} = \text{Molarity} \times \text{Liters of solution}\) \(\implies 0.0261 \text{ mol} = 0.8415 \text{ M} \times \text{Liters of solution}\) \(\implies \text{Liters of solution} = \frac{0.0261 \text{ mol}}{0.8415 \text{ M}} = 0.0310 \text{ L}\) \(\implies \text{Volume of Solution} = 31.0 \text{ mL}\)

Calculate the mass of solute for (b)

Using the formula for molarity and the volume of solution, we can find the mass of solute in (b). For (b): \(\text{Moles of solute} = \text{Molarity} \times \text{Liters of solution} = 0.299 \text{ M} \times 283 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.0846 \text{ mol}\) Mass of solute = Moles of solute \(\times\) Molar mass \(\implies \text{Mass of solute} = 0.0846 \text{ mol} \times 159.61 \text{ g/mol} = 13.51 \text{ g}\) Now the completed table is: $$ \begin{array}{cccc} \hline & \text { Mass of Solute } & \text { Volume of Solution } & \text{ Molarity } \\ \hline \text { (a) } & 12.50 \mathrm{~g} & 478 \mathrm{~mL} & 0.1638 \mathrm{M} \\ \text { (b) } & 13.51 \mathrm{~g} & 283 \mathrm{~mL} & 0.299 \mathrm{M} \\ \text { (c) } & 4.163 \mathrm{~g} & 31.0 \mathrm{~mL} & 0.8415 \mathrm{M} \\ \hline \end{array} $$