What volume of `C Cl_4` (d = 1.6 g/cc) contain 6.02 x 10^(25) `C Cl_4` molecules (CI = 35.5)

Options

A. 10.5 L
B. 250 mL
C. 9.625 L
D. 1.712 L

Answer

To determine the volume of carbon tetrachloride (CCl₄) that contains \(6.02 \times 10^{25}\) molecules, we can follow these steps:

### Step 1: Calculate the number of moles of CCl₄
We know that one mole of any substance contains \(6.022 \times 10^{23}\) molecules (Avogadro's number). Therefore, we can find the number of moles of CCl₄ in \(6.02 \times 10^{25}\) molecules using the formula:

\[ \text{Number of moles} = \frac{\text{Number of molecules}}{\text{Avogadro's number}} \]

Substituting the values:

\[ \text{Number of moles} = \frac{6.02 \times 10^{25}}{6.022 \times 10^{23}} \approx 100 \text{ moles} \]

### Step 2: Calculate the molar mass of CCl₄
The molar mass of CCl₄ can be calculated by adding the atomic masses of its constituent elements:

- Carbon (C) = 12 g/mol
- Chlorine (Cl) = 35.5 g/mol (there are 4 Cl atoms)

Thus, the molar mass of CCl₄ is:

\[ \text{Molar mass of CCl₄} = 12 + 4 \times 35.5 = 12 + 142 = 154 \text{ g/mol} \]

### Step 3: Calculate the total mass of CCl₄
Now, we can calculate the total mass of CCl₄ using the number of moles and the molar mass:

\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]

Substituting the values:

\[ \text{Mass} = 100 \text{ moles} \times 154 \text{ g/mol} = 15400 \text{ g} \]

### Step 4: Calculate the volume of CCl₄ using its density
We know that density (d) is defined as mass (m) divided by volume (V):

\[ d = \frac{m}{V} \implies V = \frac{m}{d} \]

Substituting the values:

\[ V = \frac{15400 \text{ g}}{1.6 \text{ g/cm}^3} = 9625 \text{ cm}^3 \]

### Step 5: Convert the volume from cm³ to liters
Since \(1 \text{ cm}^3 = 0.001 \text{ L}\), we convert the volume:

\[ V = 9625 \text{ cm}^3 \times 0.001 \text{ L/cm}^3 = 9.625 \text{ L} \]

### Final Answer
The volume of CCl₄ that contains \(6.02 \times 10^{25}\) molecules is approximately **9.625 liters**.

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