How many bromine atoms are present in 37.9 g of ch2br2?

The answer to your question is: 6.55 x 10 ²³ atoms of Br

Explanation:

CH2Br2 = 37.9 g

MW CH2Br2 = (12 x 1) + (2 x 1) + (80 x 2) = 174 g

                   174 g of CH2Br2  160 g of Br2

                   37.9 g of CH2Br2       x

                x = 37.9 x 160/174 = 34.85 g of Br

                      1 mol of Br   160 g Br2

                         x                  174 g Be2

               x = 174 x 1 /160 = 1.088 mol of Br2

                1 mol of Br  6.023 x 10 ²³ atoms

            1.088 mol of Br    x

                    x = 1.088 x 6.023 x 10 ²³ / 1 = 6.55 x 10 ²³ atoms

the answer is d. 74.02

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