Calculate the concentration of chloride ions in mol/dm3 when 2.22 g of CaCl2 is dissolved in 100 cm3 of water.

Here's the solution to question 6.

Step 1: Calculate the molar mass of $\text{CaCl}_2$.
The atomic mass of $\text{Ca}$ is 40 and $\text{Cl}$ is 35.5.

$$\text{Molar mass of } \text{CaCl}_2 = 40 + (2 \times 35.5) = 40 + 71 = 111 \text{ g/mol}$$

Step 2: Calculate the number of moles of $\text{CaCl}_2$.
Given mass of $\text{CaCl}_2 = 2.22 \text{ g}$.

$$\text{Moles of } \text{CaCl}_2 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{2.22 \text{ g}}{111 \text{ g/mol}} = 0.02 \text{ mol}$$

Step 3: Determine the moles of chloride ions ($\text{Cl}^-$).
One mole of $\text{CaCl}_2$ dissociates into one mole of $\text{Ca}^{2+}$ and two moles of $\text{Cl}^-$.

$$\text{CaCl}_2 \text{(aq)} \rightarrow \text{Ca}^{2+}\text{(aq)} + 2\text{Cl}^-\text{(aq)}$$

So, moles of $\text{Cl}^- = 2 \times \text{moles of } \text{CaCl}_2 = 2 \times 0.02 \text{ mol} = 0.04 \text{ mol} $$

Step 4: Convert the volume of water to $\text{dm}^3$.
Given volume of water $= 100 \text{ cm}^3$.

$$\text{Volume in } \text{dm}^3 = \frac{100 \text{ cm}^3}{1000 \text{ cm}^3/\text{dm}^3} = 0.1 \text{ dm}^3$$

Step 5: Calculate the concentration of chloride ions.

$$\text{Concentration} = \frac{\text{Moles of } \text{Cl}^-}{\text{Volume of solution in } \text{dm}^3} = \frac{0.04 \text{ mol}}{0.1 \text{ dm}^3} = 0.4 \text{ mol/dm}^3$$

The concentration of chloride ions is $\boxed{\text{0.4 mol/dm}^3}$.