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Volume occupied by one molecule of water (density $\left.=1 \mathrm{~g} \mathrm{~cm}^{-3}\right)$ is
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Verified Answer
The correct answer is:
$3.0 \times 10^{-23} \mathrm{~cm}^3$
Volume of 1 molecule of $\mathrm{H}_2 \mathrm{O}=\frac{18 \mathrm{~g}}{6.02 \times 10^{23} \times 1 \mathrm{~g} / \mathrm{cc}}$
$$
\cong 3.0 \times 10^{-23} \mathrm{~cm}^3
$$
$$
\cong 3.0 \times 10^{-23} \mathrm{~cm}^3
$$
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