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

(A) $2 \mathrm{~N}_{\mathrm{A}}$

(B) $4 \mathrm{~N}_{\mathrm{a}}$

(C) $1.2 \mathrm{~N}_{\mathrm{A}}$

(D) $2.4 \mathrm{~N}_{\mathrm{A}}$

Solution —

Moles of $D_{2} O=\frac{4}{20}=\frac{1}{5}$

No. of neutrons in $D=1$ and in oxygen $=8$

So, total no. of neutrons in 1 mole $D_{2} O=10 \mathrm{~N}_{A}$

Now, in $\frac{1}{5}$ moles $D_{2}O$, no. of neutrons $=10 \mathrm{~N}_{\mathrm{A}} \times \frac{1}{5}=2 \mathrm{~N}_{\mathrm{A}}$

So, The correct option of this question is (A)

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