Q93 / 200 · NEET 2008 (Screening) · Chemistry
Volume occupied by one molecule of water (density $=1 \mathrm{~g} \mathrm{~cm}^{-3}$ ) is
- A$9.0 \times 10^{-23} \mathrm{~cm}^3$
- B$6.023 \times 10^{-23} \mathrm{~cm}^3$
- Correct answerC$3.0 \times 10^{-23} \mathrm{~cm}^3$
- D$5.5 \times 10^{-23} \mathrm{~cm}^3$
Solution
$6.02 \times 10^{23}$ molecules of water
$\begin{aligned}
& =1 \mathrm{~mol} \\
& =18 \mathrm{~g}
\end{aligned}$
$\therefore$ Mass of one molecule of water
$\begin{aligned}
& =\frac{18}{6.023 \times 10^{23} \mathrm{~g}} \\
\because V=\frac{m}{d} & =\frac{d=\frac{m}{V}}{6.023 \times 10^{23} \times 1} \\
\therefore & \approx 3 \times 10^{-23} \mathrm{~cm}^3
\end{aligned}$
Hence, the correct answer is option: 'C'.