A piece of copper weighs 0.635 g. How many atoms of copper does it contain?

6.02 \times 10^{- 19}

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6.02 \times 10^{19}

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6.02 \times 10^{- 21}

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6.02 \times 10^{21}

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Solution

The correct option is D $6.02 \times 10^{21}$
no. of moles = $\frac{0.635}{63.5}$ = 0.01
1 mole contains 6.02x 10 $^{23}$ atoms of copper
Therefore 0.01 mol contains 0.01x 6.02 x10 $^{23}$ = 6.02 x10 $^{21}$

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