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

Explanation

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 \(1 \times 10^{25}\) Cl atoms, we can follow these steps:

### Step 1: Determine the number of moles of Cl atoms needed.
Since each molecule of CCl₄ contains 4 Cl atoms, we can calculate the number of moles of Cl atoms required.

\[ \text{Number of moles of Cl atoms} = \frac{1 \times 10^{25} \text{ Cl atoms}}{6.022 \times 10^{23} \text{ atoms/mole}} \approx 166.1 \text{ moles of Cl} \]

### Step 2: Calculate the number of moles of CCl₄ needed.
Since 1 mole of CCl₄ produces 4 moles of Cl atoms, we can find the number of moles of CCl₄ required.

\[ \text{Moles of CCl₄} = \frac{\text{Moles of Cl}}{4} = \frac{166.1}{4} \approx 41.525 \text{ moles of CCl₄} \]

### Step 3: Calculate the mass of CCl₄ required.
The molar mass of CCl₄ can be calculated as follows:

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

Now, we can calculate the total mass of CCl₄ needed:

\[ \text{Mass of CCl₄} = \text{Moles of CCl₄} \times \text{Molar mass of CCl₄} = 41.525 \text{ moles} \times 154 \text{ g/mole} \approx 639.2 \text{ g} \]

### Step 4: Convert mass to volume using density.
We know the density of CCl₄ is 1.5 g/cc. We can use this to find the volume:

\[ \text{Volume (in cc)} = \frac{\text{Mass}}{\text{Density}} = \frac{639.2 \text{ g}}{1.5 \text{ g/cc}} \approx 426.13 \text{ cc} \]

### Step 5: Convert volume from cc to liters.
Since \(1 \text{ cc} = 0.001 \text{ liters}\):

\[ \text{Volume (in liters)} = 426.13 \text{ cc} \times 0.001 \text{ liters/cc} \approx 0.426 \text{ liters} \]

### Final Answer:
Thus, the volume of liquid CCl₄ required to contain \(1 \times 10^{25}\) Cl atoms is approximately **0.426 liters**.

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