Math Snap

STEP 1

1. The solubility of $\mathrm{CdWO}_{4}$ is $0.4633 \, \mathrm{g/L}$ at $20^{\circ} \mathrm{C}$.
2. A solution is unsaturated if the concentration is less than the solubility, saturated if it equals the solubility, and supersaturated if it exceeds the solubility.

STEP 2

1. Calculate the concentration of each solution in $\mathrm{g/L}$.
2. Compare each concentration to the solubility to categorize the solutions.

STEP 3

Calculate the concentration of Solution 1:
Given: $9.018 \times 10^{-2} \, \mathrm{g}$ in $209.6 \, \mathrm{mL}$.
Convert mL to L:
$$209.6 \, \mathrm{mL} = 0.2096 \, \mathrm{L}$$ Calculate concentration:
$$\text{Concentration} = \frac{9.018 \times 10^{-2} \, \mathrm{g}}{0.2096 \, \mathrm{L}} = 0.4303 \, \mathrm{g/L}$$

STEP 4

Calculate the concentration of Solution 2:
Given: $4.814 \times 10^{-3} \, \mathrm{g}$ in $10.39 \, \mathrm{mL}$.
Convert mL to L:
$$10.39 \, \mathrm{mL} = 0.01039 \, \mathrm{L}$$ Calculate concentration:
$$\text{Concentration} = \frac{4.814 \times 10^{-3} \, \mathrm{g}}{0.01039 \, \mathrm{L}} = 0.4632 \, \mathrm{g/L}$$

STEP 5

Calculate the concentration of Solution 3:
Given: $5.962 \times 10^{-2} \, \mathrm{g}$ in $117.4 \, \mathrm{mL}$.
Convert mL to L:
$$117.4 \, \mathrm{mL} = 0.1174 \, \mathrm{L}$$ Calculate concentration:
$$\text{Concentration} = \frac{5.962 \times 10^{-2} \, \mathrm{g}}{0.1174 \, \mathrm{L}} = 0.5080 \, \mathrm{g/L}$$

STEP 6

Calculate the concentration of Solution 4:
Given: $3.117 \times 10^{-1} \, \mathrm{g}$ in $651.2 \, \mathrm{mL}$.
Convert mL to L:
$$651.2 \, \mathrm{mL} = 0.6512 \, \mathrm{L}$$ Calculate concentration:
$$\text{Concentration} = \frac{3.117 \times 10^{-1} \, \mathrm{g}}{0.6512 \, \mathrm{L}} = 0.4786 \, \mathrm{g/L}$$

Compare concentrations to solubility:
- Solution 1: $0.4303 \, \mathrm{g/L} < 0.4633 \, \mathrm{g/L}$ (Unsaturated)
- Solution 2: $0.4632 \, \mathrm{g/L} \approx 0.4633 \, \mathrm{g/L}$ (Saturated)
- Solution 3: $0.5080 \, \mathrm{g/L} > 0.4633 \, \mathrm{g/L}$ (Supersaturated)
- Solution 4: $0.4786 \, \mathrm{g/L} > 0.4633 \, \mathrm{g/L}$ (Supersaturated)
The categorization is as follows:
- Solution 1: Unsaturated
- Solution 2: Saturated
- Solution 3: Supersaturated
- Solution 4: Supersaturated