The question asks how many cubic centimeters (cm³) of a 0.06 N (Normal) solution can be prepared using 6.84 grams of Al₂(SO₄)₃.
Answer:
First, let’s calculate the molar mass of Al₂(SO₄)₃:
- Aluminum (Al): (27 , \text{g/mol} \times 2 = 54 , \text{g/mol})
- Sulfur (S): (32 , \text{g/mol} \times 3 = 96 , \text{g/mol})
- Oxygen (O): (16 , \text{g/mol} \times 12 = 192 , \text{g/mol})
Total molar mass of Al₂(SO₄)₃ = (54 + 96 + 192 = 342 , \text{g/mol})
Next, calculate the moles of Al₂(SO₄)₃ in 6.84 grams:
The normality (N) of a solution relates to the molarity (M) through the equivalent factor, which for Al₂(SO₄)₃, is 6, because Al₂(SO₄)₃ provides 6 equivalents of sulfate ions (SO₄²⁻) per mole:
[ \text{Normality} = \text{Molarity} \times \text{n} ]
For a 0.06 N solution:
[ 0.06 , \text{eq/L} = \text{Molarity} \times 6 ]
Thus, the molarity is:
[ \text{Molarity} = \frac{0.06}{6} = 0.01 , \text{mol/L} ]
From the moles calculated, the volume ( V ) in liters can be found using:
Convert liters to cubic centimeters:
[ 2 , \text{L} = 2000 , \text{cm}^3 ]
Final Answer:
2000 cm³