How many litres of liquid `C Cl_(4)` (d=1.5 g/cc) must be measured out to contain `1 xx 10^(25) C Cl_(4)` molecules?

Answer

To solve the problem of how many liters of liquid CCl₄ (with a density of 1.5 g/cc) must be measured out to contain \(1 \times 10^{25}\) CCl₄ molecules, we can follow these steps:

### Step 1: Calculate the number of moles of CCl₄
To find the number of moles (n) from the number of molecules (N), we use Avogadro's number (\(6.022 \times 10^{23}\) molecules/mol):

\[ n = \frac{N}{N_A} \]

Where:
- \(N = 1 \times 10^{25}\) molecules
- \(N_A = 6.022 \times 10^{23}\) molecules/mol

Substituting the values:

\[ n = \frac{1 \times 10^{25}}{6.022 \times 10^{23}} \approx 16.61 \text{ moles} \]

### Step 2: Calculate the mass of CCl₄ required
Next, we need to find the mass of CCl₄ using the molar mass. The molar mass of CCl₄ can be calculated as follows:

- Carbon (C) = 12.01 g/mol
- Chlorine (Cl) = 35.45 g/mol

Thus, the molar mass of CCl₄ is:

\[ \text{Molar mass of CCl}_4 = 12.01 + (4 \times 35.45) = 12.01 + 141.8 = 153.81 \text{ g/mol} \]

Now, we can calculate the mass (m) using the formula:

\[ m = n \times \text{Molar mass} \]

Substituting the values:

\[ m = 16.61 \text{ moles} \times 153.81 \text{ g/mol} \approx 2557.7 \text{ g} \]

### Step 3: Convert mass to volume using density
Now we can find the volume (V) of CCl₄ using its density (d):

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

Where:
- \(d = 1.5 \text{ g/cc} = 1.5 \text{ g/mL}\)
- \(m = 2557.7 \text{ g}\)

Substituting the values:

\[ V = \frac{2557.7 \text{ g}}{1.5 \text{ g/mL}} \approx 1705.13 \text{ mL} \]

### Step 4: Convert volume from mL to liters
Since 1 L = 1000 mL, we convert the volume to liters:

\[ V \text{ (in liters)} = \frac{1705.13 \text{ mL}}{1000} \approx 1.705 \text{ L} \]

### Final Answer
Thus, the volume of liquid CCl₄ required is approximately **1.705 liters**.

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