The weight of `1 xx 10^22` molecules of `CuSO_4. 5H_2O` is:

Answer

To find the weight of \( 1 \times 10^{22} \) molecules of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \), we can follow these steps:

### Step 1: Calculate the number of moles
We will use Avogadro's number, which is \( 6.022 \times 10^{23} \) molecules per mole, to convert the number of molecules to moles.

\[ \text{Number of moles} = \frac{\text{Number of molecules}}{\text{Avogadro's number}} = \frac{1 \times 10^{22}}{6.022 \times 10^{23}} \]

Calculating this gives:

\[ \text{Number of moles} = \frac{1 \times 10^{22}}{6.022 \times 10^{23}} \approx 0.0166 \text{ moles} \]

### Step 2: Calculate the molar mass of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \)
We need to find the molar mass of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \) by adding the atomic masses of its components:

- Copper (Cu): \( 63.5 \, \text{g/mol} \)
- Sulfur (S): \( 32.0 \, \text{g/mol} \)
- Oxygen (O): \( 16.0 \, \text{g/mol} \) (4 atoms in \( \text{CuSO}_4 \))
- Water (H2O): \( 18.0 \, \text{g/mol} \) (5 molecules of water)

Calculating the total molar mass:

\[ \text{Molar mass of } \text{CuSO}_4 \cdot 5\text{H}_2\text{O} = 63.5 + 32 + (4 \times 16) + (5 \times 18) \]
\[ = 63.5 + 32 + 64 + 90 = 249.5 \, \text{g/mol} \]

### Step 3: Calculate the mass
Now we can use the number of moles and the molar mass to find the mass:

\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]
\[ \text{Mass} = 0.0166 \, \text{moles} \times 249.5 \, \text{g/mol} \]
\[ \text{Mass} \approx 4.14 \, \text{grams} \]

### Final Answer
The weight of \( 1 \times 10^{22} \) molecules of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \) is approximately **4.14 grams**.

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