If $N_{A}$ is Avogadro's number, then calculate the number of electrons in $4.2 g$ of $N^{3 -}$ .

2.1 N_{A}

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3 N_{A}

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6 N_{A}

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N_{A}

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Solution

The correct option is C $2.1 N_{A}$
$N^{3 -}$ contains 10 electrons.

Number of moles $= \frac{4.2}{14}$

$= 0.3 m o l e s$

No of electrons in $0.3 m o l e s = 0.3 \times N_{A} \times 10$

$= 3 N_{A}$

Hence, option B is correct.

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