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Q.

The number of electrons in $3.1 mg {NO}_{3}^{-}$ is:

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a

$1.6 \times 10^{- 3}$

b

32

c

$9.6 \times 10^{20}$

d

$9.6 \times 10^{23}$

answer is C.

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Detailed Solution

The number of electrons in this case;

$\frac{3.1 \times 10^{- 3}}{62} \times 6.023 \times 10^{23} \times 32$

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