What volume of carbon tetrachloride (density = 2.71 g/mL) contains 2.46x10^25 chlorine atoms?

one C per molecule?

6.02 * 10^23 molecules per mol

2.46 * 10^25 molecules *(1/6.02*10^23)
= 40.9 mols

How many grams?
40.9 * (12 + 4(35.5) )

what volume?
grams / 2.71 g/mL

but how do you get the right amount of mL?

I did: 2.46x10^25 Cl atoms x 1 atom/4 atoms x 1 mol/6.02x10^23 x 154 g/1 mol x 1 mL/2.71 g

And the correct answer is apparently 572 mL when I got 581 mL.

To find the volume of carbon tetrachloride containing a specific number of chlorine atoms, we can use the concept of molar volume and Avogadro's number.

First, we need to calculate the molar mass of carbon tetrachloride (CCl4). Carbon has an atomic mass of 12.01 g/mol, and chlorine has an atomic mass of 35.45 g/mol. Since there are four chlorine atoms, the molar mass of carbon tetrachloride is:

Molar mass of CCl4 = (4 * 35.45 g/mol) + 12.01 g/mol = 153.82 g/mol

Next, we can use Avogadro's number (6.022 × 10^23) to find the number of moles in 2.46 × 10^25 chlorine atoms:

Number of moles = (2.46 × 10^25 atoms) / (6.022 × 10^23 atoms/mol) ≈ 40.88 mol

Now, we can find the mass of the carbon tetrachloride corresponding to this number of moles:

Mass = Number of moles × Molar mass
= 40.88 mol × 153.82 g/mol
= 6,296.85 g

Finally, we can use the density of carbon tetrachloride to find the volume:

Density = Mass / Volume

Rearranging the equation:

Volume = Mass / Density
= 6,296.85 g / 2.71 g/mL
≈ 2,323.99 mL or 2.32 L (rounded to two decimal places)

Therefore, approximately 2.32 liters of carbon tetrachloride would contain 2.46 × 10^25 chlorine atoms.