Volume occupied by one molecule of water (density $=1 \mathrm{~g} \mathrm{~cm}^{-3}$ ) is
- $9.0 \times 10^{-23} \mathrm{~cm}^3$
- $6.023 \times 10^{-23} \mathrm{~cm}^3$
- $3.0 \times 10^{-23} \mathrm{~cm}^3$
- $5.5 \times 10^{-23} \mathrm{~cm}^3$
Solution
$\begin{aligned}
& =1 \mathrm{~mol} \\
& =18 \mathrm{~g}
\end{aligned}$
$\therefore$ Mass of one molecule of water
$\begin{aligned}
& =\frac{18}{6.023 \times 10^{23} \mathrm{~g}} \\
\because V=\frac{m}{d} & =\frac{d=\frac{m}{V}}{6.023 \times 10^{23} \times 1} \\
\therefore & \approx 3 \times 10^{-23} \mathrm{~cm}^3
\end{aligned}$
Asked in: NEET 2008 (Screening)