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

Options

A. `4.14 g`
B. `5.14 g`
C. `6.14 g`
D. `7.14 g`

Explanation

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

### Step 1: Determine the molecular weight of \( CuSO_4 \cdot 5H_2O \)

1. **Calculate the molecular weight of each component:**
- Copper (Cu): 63.5 g/mol
- Sulfur (S): 32.1 g/mol
- Oxygen (O): 16.0 g/mol (4 from \( SO_4 \) and 10 from \( 5H_2O \))
- Hydrogen (H): 1.0 g/mol (10 from \( 5H_2O \))

2. **Calculate the total molecular weight:**
\[ \text{Molecular weight of } CuSO_4 \cdot 5H_2O = 63.5 + 32.1 + (4 \times 16.0) + (10 \times 1.0) \]
\[ = 63.5 + 32.1 + 64.0 + 10.0 = 169.6 \text{ g/mol} \]

### Step 2: Use Avogadro's number to find the weight of \( 1 \times 10^{22} \) molecules

1. **Use Avogadro's number (approximately \( 6.022 \times 10^{23} \) molecules/mol) to find the fraction of a mole:**
\[ \text{Number of moles} = \frac{1 \times 10^{22}}{6.022 \times 10^{23}} \approx 0.0166 \text{ moles} \]

### Step 3: Calculate the weight of \( 1 \times 10^{22} \) molecules

1. **Calculate the weight using the molecular weight:**
\[ \text{Weight} = \text{Number of moles} \times \text{Molecular weight} \]
\[ = 0.0166 \text{ moles} \times 249 \text{ g/mol} \approx 4.14 \text{ grams} \]

### Final Answer

The weight of \( 1 \times 10^{22} \) molecules of \( CuSO_4 \cdot 5H_2O \) is approximately **4.14 grams**.

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