The number of water molecules present in a drop of water (Volume $0.0018ml$) at room temperature is

The correct option is A $6.023\times {10}^{19}$
As we know, density is equal to mass upon volume.
We know density of water = $1gm{l}^{-1}$. Hence for water 1 g = 1 ml and 0.0018 ml = 0.0018 g.
Number of moles =$\frac{\text{Weight}}{\text{Molecular weight}}$
$=\frac{0.0018}{18}$
$=1\times {10}^{-4}mol$
$\therefore$ No. of water molecules = Moles $\times N_A$$=6.023\times {10}^{23}\times 1\times {10}^{-4}$
$=6.023\times {10}^{19}$ molecules.