`1.245g` of `CuSO_(4).xH_(2)O` was dissolved in water and `H_(2)S` was passed until CuS was completely precipitated. The `H_(2)SO_(4)` Produced in the filtrated required 10 ml of 1M NaOH solution . Calculate X

Answer

To solve the problem, we need to determine the value of \( x \) in the compound \( CuSO_4 \cdot xH_2O \) based on the given data. Let's break it down step by step.

### Step 1: Determine the moles of \( CuSO_4 \cdot xH_2O \)

1. **Calculate the molar mass of \( CuSO_4 \)**:
- Copper (Cu) = 63.5 g/mol
- Sulfur (S) = 32.1 g/mol
- Oxygen (O) = 16 g/mol (4 O atoms)
- Molar mass of \( CuSO_4 = 63.5 + 32.1 + (4 \times 16) = 159.6 \, g/mol \)

2. **Let the molar mass of \( xH_2O \)** be \( 18x \, g/mol \) (since the molar mass of water \( H_2O \) is 18 g/mol).

3. **Total molar mass of \( CuSO_4 \cdot xH_2O \)**:
\[ \text{Molar mass} = 159.6 + 18x \]

4. **Calculate the moles of \( CuSO_4 \cdot xH_2O \)**:
\[ \text{Moles of } CuSO_4 \cdot xH_2O = \frac{\text{mass}}{\text{molar mass}} = \frac{1.245}{159.6 + 18x} \]

### Step 2: Determine the moles of \( H_2SO_4 \) produced

1. **From the problem, we know that 10 mL of 1 M NaOH is used**:
- Moles of NaOH = Volume (L) × Molarity (mol/L) = \( 0.010 \, L \times 1 \, mol/L = 0.01 \, mol \)

2. **Using the balanced reaction**:
\[ H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O \]
- This shows that 1 mole of \( H_2SO_4 \) reacts with 2 moles of \( NaOH \).
- Therefore, moles of \( H_2SO_4 \) produced:
\[ \text{Moles of } H_2SO_4 = \frac{0.01}{2} = 0.005 \, mol \]

### Step 3: Set up the equation

1. **From the stoichiometry of the reaction**:
\[ \text{Moles of } H_2SO_4 = \text{Moles of } CuSO_4 \cdot xH_2O \]
- Thus, we can set up the equation:
\[ 0.005 = \frac{1.245}{159.6 + 18x} \]

### Step 4: Solve for \( x \)

1. **Cross-multiply to solve for \( x \)**:
\[ 0.005(159.6 + 18x) = 1.245 \]
\[ 0.798 + 0.09x = 1.245 \]
\[ 0.09x = 1.245 - 0.798 \]
\[ 0.09x = 0.447 \]
\[ x = \frac{0.447}{0.09} \approx 4.967 \]

### Step 5: Conclusion

Since \( x \) is approximately 4.967, we round it to the nearest whole number:
\[ x \approx 5 \]

Thus, the value of \( x \) is 5, and the formula of the hydrated copper sulfate is \( CuSO_4 \cdot 5H_2O \).