Ordinary water contains one part of heavy water per 6000 parts by weight. The number of heavy water molecules present in a drop of water of volume $0.01 mL$ is (density of water is $1 g/mL$ )

Heavy water is $D_{2}O$ and ordinary water is $H_{2}O$ respectively.

$0.01 mL$ of water $=0.01 g$

Molar mass of $D_{2}O=2 g$

Now, $6000 g$ of $H_{2}O=1 g$ of $D_{2}O$

$1 g$ of $H_{2}O=1/6000 g{ofD}_{2}O$

So, $0.01 g$ of $H_{2}O=1/6000\times 0.01=0.16\times {10}^{-5} g$ of $D_{2}O$

Now,

No. of moles of $D_{2}O=$ given mass/molar mass

$=0.16\times {10}^{-5}/20=0.008\times {10}^{-5}$

1 mol. of $D_{2}O=6.022\times {10}^{23}$

$$\begin{gathered} 0.008\times {10}^{-5} moles contains =6.022\times {10}^{23}\times 0.008\times {10}^{-5} \\ =4.8\times {10}^{16} or5\times {10}^{16} \end{gathered}$$

Therefore, option (C) is correct.