What volume of carbon tetrachloride (density = 2.71 g/mL) contains 2.46x10^25 chlorine atoms?
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 calculate the volume of carbon tetrachloride containing a given number of chlorine atoms, you need to use the following steps:
Step 1: Calculate the number of moles of chlorine atoms using Avogadro's number.
Number of moles of chlorine = (2.46x10^25 Cl atoms) / (6.02x10^23 atoms/mol) = 40.864 mol
Step 2: Use the molar mass of carbon tetrachloride to convert moles of chlorine to grams.
Molar mass of carbon tetrachloride (CCl4) = 12.01 g/mol (carbon) + (4 x 35.45 g/mol) (chlorine) = 153.82 g/mol
Mass of chlorine = 40.864 mol x 153.82 g/mol = 6285.22 g
Step 3: Convert grams of chlorine to volume of carbon tetrachloride using the density.
Volume of carbon tetrachloride = (6285.22 g) / (2.71 g/mL) = 2316.94 mL
The correct answer, rounded to the nearest mL, would be 2317 mL. Therefore, it seems that there might have been a slight rounding error in your calculations that caused the discrepancy.