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Q.

Number of electrons present in 5.1 grams of ammonia are( $N_{A} = 6.023 \times 10^{23}$ )

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a

3 $N_{A}$

b

6.022 $N_{A}$

c

5.1 $N_{A}$

d

6 $N_{A}$

answer is A.

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Detailed Solution

$Number of moles of {NH}_{3} = \frac{5.1}{17} = 0.3 Number of electrons = 0.3 \times 10 \times N_{A} = 3 N_{A}$

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