Q11 / 150 · MHT CET 2024 (09 May Shift 1) · Chemistry
What is the volume occupied by 1 molecule of water, if its density is $1 \mathrm{~g} \mathrm{~cm}^{-3}$ ?
- A$9.0 \times 10^{-23} \mathrm{~cm}^3$
- Correct answerB$\quad 2.98 \times 10^{-23} \mathrm{~cm}^3$
- C$6.023 \times 10^{-23} \mathrm{~cm}^3$
- D$5.50 \times 10^{-23} \mathrm{~cm}^3$
Solution
Mass of $6.022 \times 10^{23}$ molecules of water $=1.8 \mathrm{~g}$
$\therefore \quad$ Mass of 1 molecule of water
$\begin{aligned}
& =\frac{18 \mathrm{~g} \times 1}{6.022 \times 10^{23}}=2.98 \times 10^{-23} \mathrm{~g} \\
& \text { Density }=\frac{\text { mass }}{\text { volume }}
\end{aligned}$
$\therefore \quad$ Volume occupied by 1 molecule of water
$=\frac{\text { mass }}{\text { density }}=\frac{2.98 \times 10^{-23} \mathrm{~g}}{1 \mathrm{~g} \mathrm{~cm}^{-3}}=2.98 \times 10^{-23} \mathrm{~cm}^3$
Hence, the correct answer is option: 'B'.