State two methods that can be used to remove water hardness. (b) Define the term solubility. Use the data to calculate the solubility of solid Q.

$$\text{Mass of saturated solution} = \text{Mass of (dish + saturated solution)} - \text{Mass of empty dish}$$ $$\text{Mass of saturated solution} = 62.5 \text{ g} - 46.5 \text{ g} = 16.0 \text{ g}$$

$$\text{Mass of dry solid Q} = \text{Mass of (dish + dry solid Q)} - \text{Mass of empty dish}$$ $$\text{Mass of dry solid Q} = 50.4 \text{ g} - 46.5 \text{ g} = 3.9 \text{ g}$$

$$\text{Mass of water} = \text{Mass of saturated solution} - \text{Mass of dry solid Q}$$ $$\text{Mass of water} = 16.0 \text{ g} - 3.9 \text{ g} = 12.1 \text{ g}$$

$$\text{Solubility} = \frac{\text{Mass of solid Q}}{\text{Mass of water}} \times 100 \text{ g}$$ $$\text{Solubility} = \frac{3.9 \text{ g}}{12.1 \text{ g}} \times 100 \text{ g}$$ $$\text{Solubility} = 32.2314... \text{ g/100 g water}$$

The solubility of solid Q is $\boxed{32.2 \text{ g/100 g water}}$.

8. (a) What is meant by lattice energy?
Lattice energy is the energy change when one mole of an ionic compound is formed from its gaseous ions under standard conditions.

(b) Use the information below to calculate the heat of solution of calcium chloride.
The heat of solution ($\Delta H_{sol}$) can be calculated using Hess's Law:

$$\Delta H_{sol} = \text{Lattice energy (dissociation)} + \text{Hydration energy of cations} + \text{Hydration energy of anions}$$

The given lattice energy for $\text{CaCl}_2$ is -2195 kJ/mol, which is the energy released when the lattice is formed from gaseous ions. Therefore, the energy required to dissociate the lattice into gaseous ions is $+2195 \text{ kJ/mol}$.
For $\text{CaCl}_2$, there is one $\text{Ca}^{2+}$ ion and two $\text{Cl}^-$ ions.

$$\Delta H_{sol} = -(-2195 \text{ kJ/mol}) + (-189 \text{ kJ/mol}) + (2 \times -384 \text{ kJ/mol})$$ $$\Delta H_{sol} = 2195 \text{ kJ/mol} - 189 \text{ kJ/mol} - 768 \text{ kJ/mol}$$

$$\Delta H_{sol} = 2195 - 189 - 768$$ $$\Delta H_{sol} = 1238 \text{ kJ/mol}$$

The heat of solution of calcium chloride is $\boxed{+1238 \text{ kJ/mol}}$.

9. (a) State Boyle's law.
Boyle's law states that for a fixed mass of gas at constant temperature, the pressure of the gas is inversely proportional to its volume.

(b) A gas occupies 500 $\text{cm}^3$ at 37°C and 100,000 Pa. What will be its volume at 10°C and 101,325 Pa?
Use the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$

$$T_1 = 37^\circ\text{C} + 273.15 = 310.15 \text{ K}$$ $$T_2 = 10^\circ\text{C} + 273.15 = 283.15 \text{ K}$$

$$P_1 = 100,000 \text{ Pa}$$ $$V_1 = 500 \text{ cm}^3$$ $$T_1 = 310.15 \text{ K}$$ $$P_2 = 101,325 \text{ Pa}$$ $$T_2 = 283.15 \text{ K}$$ $$V_2 = ?$$

$$V_2 = \frac{(100,000 \text{ Pa}) \times (500 \text{ cm}^3) \times (283.15 \text{ K})}{(101,325 \text{ Pa}) \times (310.15 \text{ K})}$$ $$V_2 = \frac{14157500000}{31499993.75}$$ $$V_2 = 449.43 \text{ cm}^3$$

The volume of the gas will be $\boxed{449 \text{ cm}^3}$ (to 3 significant figures).

10. Describe how a solid sample of copper (II) carbonate can be prepared starting with copper metal.

1. React the copper metal with concentrated nitric acid to form a soluble copper (II) nitrate solution. $\text{Cu (s)} + 4\text{HNO}_3 \text{(conc)} \rightarrow \text{Cu(NO}_3)_2 \text{(aq)} + 2\text{NO}_2 \text{(g)} + 2\text{H}_2\text{O (l)}$
2. Add a solution of a soluble carbonate, such as sodium carbonate, to the copper (II) nitrate solution. This will cause copper (II) carbonate to precipitate as it is insoluble. $\text{Cu(NO}_3)_2 \text{(aq)} + \text{Na}_2\text{CO}_3 \text{(aq)} \rightarrow \text{CuCO}_3 \text{(s)} + 2\text{NaNO}_3 \text{(aq)}$
3. Filter the precipitate (copper (II) carbonate) from the solution.
4. Wash the solid copper (II) carbonate with distilled water to remove any soluble impurities.
5. Dry the solid copper (II) carbonate, for example, by placing it in a desiccator or in a warm oven.