SOLUTION : Since density of water $=1 \cdot 0 g / mL$
$\therefore \quad$ Mass of onc drop of watcr $=$ volume $\times$ density $=0.05 \times 1 g =0.05 g$ No. of mole of water $=\frac{\text { mass }}{\text { mol. wt. }}=\frac{0 \cdot 05}{18}$ mole.
Hence, no. of molecules in one drop of water
$=\frac{0 \cdot 05}{18} \times 6.02 \times 10^{23} \text { molecules }= 1 . 6 7 \times 10^{21} \text { molecules. }$