The number of water molecules present in a drop of water (volume $0.0018 \mathrm{~mL}$ ) density $=1…
The number of water molecules present in a drop of water (volume $0.0018 \mathrm{~mL}$ ) density $=1 \mathrm{g}\mathrm{~mL}^{-1}$ at room temperature is
$1.084 \times 10^{18}$
$6.023 \times 10^{19}$
$4.84 \times 10^{17}$
$6.023 \times 10^{23}$
Solution
As we know that,
Density $=\frac{\text { mass }}{\text { volume }}$
$\Rightarrow$ mass $=$ density $\times$ volume
Density of water $=1 \mathrm{~g} / \mathrm{ml}$
Volume of water $=0.0018 \mathrm{~mL}$ (Given)
$\therefore$ Mass of $0.0018 \mathrm{~mL}$ water $=1 \times 0.0018=0.0018 \mathrm{gm}$
Molar mass of water $=18 \mathrm{gm} / \mathrm{mol}$
Number of moles in $0.0018 \mathrm{gm}$ of water $=\frac{\text { Given mass }}{\text { Molar mass }}=\frac{0.0018}{18}=0.0001 \mathrm{~mole}$
As we know that,
Number of molecules of water in 1 mole $=6.022 \times 10^{23}$ molecules
$\therefore$ Number of molecules of water in $0.0001$ mole $=6.022 \times 10^{3} \times 0.0001=$ $6.023 \times 10^{19}$
Hence, the number of water molecules present in a drop of water (volume $=0.0018 \mathrm{ml}$) at room temperature are $6.023 \times 10^{19}$.