Question

Determine the approximate weight of $1.204\times {10}^{24}$ bromine atoms?

• A. 80 grams
• B. 120 grams
• C. 160 grams
• D. 180 grams
• E. 200 grams

Answer 1

Answer: (C)

160 grams

1 mole $(6.022\times {10}^{23})$ atoms of Bromine has $80g$ mass.
So, $1.204\times {10}^{24}$ atoms has $\frac{80}{6.022\times {10}^{23}}\times 1.204\times {10}^{24}=159.9grams$.