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Q.

Total number of neutrons present in 4g of heavy water $(D_{2} O)$ is : (Where $N_{A}$ represents Avogadro's number)

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a

$1.2 N_{A}$

b

$2 N_{A}$

c

$4 N_{A}$

d

$2.4 N_{A}$

answer is D.

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Detailed Solution

Total number of neutrons preset,

$\begin{array}{l} = \frac{4}{20} \times 10 \times N_{A} \\ = 2 N_{A} \end{array}$

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