How many electrons are present in $2 \times 10^{- 3} {moles of}_{8}^{18} O^{- 2}$

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$1 .21 \times 10^{21}$

$1.2 \times 10^{22}$

$9 .6 \times 10^{21}$

$1 .9 \times 10^{22}$

answer is C.

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Detailed Solution

Number of $\bar{e} in O^{- 2} = 8 + 2 = 10$
Number of ${\bar{es}}^{-} as 2 \times 10^{- 3} moles O^{- 2}$ of = $2 \times 10^{- 3} \times 6 \times 10^{23} \times 10 = 1 .2 \times 10^{22}$

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