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How many grams are in 1.19x10^27 particles of cl2

How many grams are in 1.19x10^27 particles of cl2-example-1
User Evgeniuz
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1 Answer

5 votes

Answer:

Option D. 1.40x10^5g

Step-by-step explanation:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02x10^23 particles. This equally means that 1mole of Cl2 contains 6.02x10^23 particles.

1 mole of Cl2 =2 x 35.5 = 71g.

Now, If 71g of Cl2 contains 6.02x10^23 particles,

Then Xg of Cl2 will contain 1.19x10^27 particles i.e

Xg of Cl2 = (71x1.19x10^27)/(6.02x10^23)

Xg of Cl2 = 1.40x10^5g

Therefore, 1.40x10^5g of Cl2 contain 1.19x10^27 particles

User Tchen
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