Question

what volume of CCL4 having density 1.5g/cc will contain 10^25 atoms of chlorine?


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Answer 1

Answer:

Explanation:

The question is not complete. The complete question is to find 1*10^25 atoms of Cl.  

Let 1 molecule = v*6.023 *10^23/22.4

4 atoms = v*6.023*10^23/22.4

So 1*10^25 atoms= V*6.023*10^23/22.4

Cross multiply and u will get the answer to be 0.427 L

Answer 2

One CCl4 molecule contains 4 Cl atoms.

1 x 10^25 Cl atoms x (1 molecule CCl4 / 4 Cl atoms) = 2.5 x 10^24 CCl4 molecules.

Convert molecules of CCl4 to moles. One mole of CCl4 contains 6.02 x 10^23 (Avogadro's number) of CCl4 molecules.

2.5 x 10^24 CCl4 molecules x (1 mole CCl4 / 6.02 x 10^23 molecules CCl4) = 4.2 moles CCl4.

Convert moles of CCl4 to grams. The molar mass of CCl4 is equal to the sum of the atomic weights of one C and 4 Cl found on the periodic table = (1 x 12.0) + (4 x 35.45) = 153.8 g CCl4.

4.2 moles CCl4 x (153.8 g CO2 / 1 mole CO2) = 646 g CCl4.

Remember that 1 cc = 1 mL. Use the density to convert g of CCl4 to mL.

646 g CCl4 x (1 mL CCl4 / 1.5 g CCl4) = 431 mL CCl4

431 mL CO2 x (1 L CCl4 / 1000 mL CCl4) = 0.431 L

Note:-

We can round off means we can say ans as 1.5 instead of 0.431

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