What volume of` C CI_4` `(d = 1.6 g/cc)` contain ` 6.02 X 10^(25)` `C CI_4` molecules` (CI = 35.5)`

Options

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

Answer

To solve the problem of finding the volume of CCl₄ that contains \(6.02 \times 10^{25}\) molecules, we will follow these steps:

### Step 1: Calculate the number of moles of CCl₄
To find the number of moles from the number of molecules, we use Avogadro's number, which is \(6.02 \times 10^{23}\) molecules/mol.

\[ \text{Number of moles} = \frac{\text{Number of molecules}}{\text{Avogadro's number}} = \frac{6.02 \times 10^{25}}{6.02 \times 10^{23}} = 100 \text{ moles} \]

### 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 (and there are 4 chlorine atoms)

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

### Step 3: Calculate the mass of CCl₄
Using the number of moles and the molar mass, we can find the mass:

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

### Step 4: Calculate the volume of CCl₄ using its density
We can find the volume using the formula:

\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]

Given that the density of CCl₄ is \(1.6 \text{ g/cc}\):

\[ \text{Volume} = \frac{15400 \text{ g}}{1.6 \text{ g/cc}} = 9625 \text{ cc} \]

### Step 5: Convert volume from cc to liters
Since \(1 \text{ liter} = 1000 \text{ cc}\):

\[ \text{Volume in liters} = \frac{9625 \text{ cc}}{1000} = 9.625 \text{ liters} \]

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

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