The weight of `1 xx 10^(22)` molecules of `CuSO_(4). 5H_(2)O` is ……..

Explanation

To find the weight of \( 1 \times 10^{22} \) molecules of \( CuSO_4 \cdot 5H_2O \), we can follow these steps:

### Step 1: Calculate the Molar Mass of \( CuSO_4 \cdot 5H_2O \)

1. **Identify the components**:
- Copper (Cu): 1 atom
- Sulfur (S): 1 atom
- Oxygen (O): 4 atoms from \( CuSO_4 \) and 5 from \( 5H_2O \) (total 9)
- Hydrogen (H): 10 atoms from \( 5H_2O \)

2. **Find the atomic masses**:
- Atomic mass of Cu = 63.5 g/mol
- Atomic mass of S = 32 g/mol
- Atomic mass of O = 16 g/mol
- Atomic mass of H = 1 g/mol

3. **Calculate the molar mass**:
\[ \text{Molar mass of } CuSO_4 \cdot 5H_2O = (1 \times 63.5) + (1 \times 32) + (4 \times 16) + (10 \times 1) \]
\[ = 63.5 + 32 + 64 + 10 = 169.5 \text{ g/mol} \]

### Step 2: Calculate the Mass of \( 1 \times 10^{22} \) Molecules

1. **Use Avogadro's number**:
- Avogadro's number \( N_A = 6.022 \times 10^{23} \) molecules/mol

2. **Set up the proportion**:
- The mass of 1 mole (which contains \( 6.022 \times 10^{23} \) molecules) is 169.5 g.
- Therefore, the mass of \( 1 \times 10^{22} \) molecules can be calculated using the unitary method:
\[ \text{Mass} = \left( \frac{169.5 \text{ g}}{6.022 \times 10^{23} \text{ molecules}} \right) \times (1 \times 10^{22} \text{ molecules}) \]

3. **Calculate the mass**:
\[ \text{Mass} = \frac{169.5}{6.022 \times 10^{23}} \times 1 \times 10^{22} \]
\[ = \frac{169.5 \times 1 \times 10^{22}}{6.022 \times 10^{23}} \approx 2.82 \text{ g} \]

### Final Answer
The weight of \( 1 \times 10^{22} \) molecules of \( CuSO_4 \cdot 5H_2O \) is approximately **2.82 grams**.