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Math  /  Data & Statistics

QuestionThe solubility of CdWO4\mathrm{CdWO}_{4} is 0.4633 g/LH2O0.4633 \mathrm{~g} / \mathrm{L} \mathrm{H}_{2} \mathrm{O} at 20∘C20^{\circ} \mathrm{C}. Several solutions of CdWO4\mathrm{CdWO}_{4} (at 20∘C20^{\circ} \mathrm{C} ) have been prepared. Categorize each solution as unsaturated, saturated, or supersaturated.
Unsaturated Saturated Supersaturated □\square □\square
Answer Bank
Solution 1: 9.018×10−2 g9.018 \times 10^{-2} \mathrm{~g} solute is completely dissolved in 209.6 mL water.
Solution 2: 4.814×10−3 g4.814 \times 10^{-3} \mathrm{~g} solute is completely dissolved in 10.39 mL water.
Solution 3: 5.962×10−2 g5.962 \times 10^{-2} \mathrm{~g} solute is completely dissolved in 117.4 mL water.
Solution 4: 3.117×10−1 g3.117 \times 10^{-1} \mathrm{~g} solute is completely dissolved in 651.2 mL water.

Studdy Solution

STEP 1

1. The solubility of CdWO4\mathrm{CdWO}_{4} is 0.4633 g/L0.4633 \, \mathrm{g/L} at 20∘C20^{\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 g/L\mathrm{g/L}.
2. Compare each concentration to the solubility to categorize the solutions.

STEP 3

Calculate the concentration of Solution 1:
Given: 9.018×10−2 g9.018 \times 10^{-2} \, \mathrm{g} in 209.6 mL209.6 \, \mathrm{mL}.
Convert mL to L:
209.6 mL=0.2096 L 209.6 \, \mathrm{mL} = 0.2096 \, \mathrm{L}
Calculate concentration:
Concentration=9.018×10−2 g0.2096 L=0.4303 g/L \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: 4.814×10−3 g4.814 \times 10^{-3} \, \mathrm{g} in 10.39 mL10.39 \, \mathrm{mL}.
Convert mL to L:
10.39 mL=0.01039 L 10.39 \, \mathrm{mL} = 0.01039 \, \mathrm{L}
Calculate concentration:
Concentration=4.814×10−3 g0.01039 L=0.4632 g/L \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: 5.962×10−2 g5.962 \times 10^{-2} \, \mathrm{g} in 117.4 mL117.4 \, \mathrm{mL}.
Convert mL to L:
117.4 mL=0.1174 L 117.4 \, \mathrm{mL} = 0.1174 \, \mathrm{L}
Calculate concentration:
Concentration=5.962×10−2 g0.1174 L=0.5080 g/L \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: 3.117×10−1 g3.117 \times 10^{-1} \, \mathrm{g} in 651.2 mL651.2 \, \mathrm{mL}.
Convert mL to L:
651.2 mL=0.6512 L 651.2 \, \mathrm{mL} = 0.6512 \, \mathrm{L}
Calculate concentration:
Concentration=3.117×10−1 g0.6512 L=0.4786 g/L \text{Concentration} = \frac{3.117 \times 10^{-1} \, \mathrm{g}}{0.6512 \, \mathrm{L}} = 0.4786 \, \mathrm{g/L}

STEP 7

Compare concentrations to solubility:
- Solution 1: 0.4303 g/L<0.4633 g/L0.4303 \, \mathrm{g/L} < 0.4633 \, \mathrm{g/L} (Unsaturated) - Solution 2: 0.4632 g/L≈0.4633 g/L0.4632 \, \mathrm{g/L} \approx 0.4633 \, \mathrm{g/L} (Saturated) - Solution 3: 0.5080 g/L>0.4633 g/L0.5080 \, \mathrm{g/L} > 0.4633 \, \mathrm{g/L} (Supersaturated) - Solution 4: 0.4786 g/L>0.4633 g/L0.4786 \, \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

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