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Volume occupied by one molecule of water (density = 1 g cm$^{-3}$) is :
A
$3.0 \times 10^{-23}$ cm$^3$
B
$5.5 \times 10^{-23}$ cm$^3$
C
$9.0 \times 10^{-23}$ cm$^3$
D
$6.023 \times 10^{-23}$ cm$^3$
Detailed Solution
Weight of $6.023 \times 10^{23}$ molecules (1 mole) of water = 18 g
Volume occupied by $6.023 \times 10^{23}$ molecules of water (density = 1 g cm$^{-3}$) $= \dfrac{\text{mass}}{\text{density}} = \dfrac{18\ g}{1\ g\ cm^{-3}}$
$= 18$ cm$^3$ or 18 mL
So the volume occupied by one molecule of water $= \dfrac{18}{6.023 \times 10^{23}}$ cm$^3$
$= 2.988 \times 10^{-23}$ cm$^3$
$\approx 3.0 \times 10^{-23}$ cm$^3$
Volume occupied by $6.023 \times 10^{23}$ molecules of water (density = 1 g cm$^{-3}$) $= \dfrac{\text{mass}}{\text{density}} = \dfrac{18\ g}{1\ g\ cm^{-3}}$
$= 18$ cm$^3$ or 18 mL
So the volume occupied by one molecule of water $= \dfrac{18}{6.023 \times 10^{23}}$ cm$^3$
$= 2.988 \times 10^{-23}$ cm$^3$
$\approx 3.0 \times 10^{-23}$ cm$^3$