Volume occupied by one molecule of water (den | vedclass

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(A) The molar mass of water $(H_2O)$ is $18.02 \ g \ mol^{-1}$.Given density of water is $1 \ g \ cm^{-3}$.Volume of $1 \ mol$ of water $= 	ext{Mass}

Vedclass · Accepted Answer

$3.0$

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(A) The molar mass of water $(H_2O)$ is $18.02 \ g \ mol^{-1}$.Given density of water is $1 \ g \ cm^{-3}$.Volume of $1 \ mol$ of water $= 	ext{Mass} / 	ext{Density} = 18.02 \ g / 1 \ g \ cm^{-3} = 18.02 \ cm^3$.Since $1 \ mol$ contains $6.022 	imes 10^{23}$ molecules,the volume occupied by one molecule is:$V = 18.02 \ cm^3 / 6.022 	imes 10^{23} 	ext{ molecules} \approx 2.99 	imes 10^{-23} \ cm^3 \approx 3.0 	imes 10^{-23} \ cm^3$.

Volume occupied by one molecule of water (density $= 1 \ g \ cm^{-3}$) is ...... $ \times 10^{-23} \ cm^3$.

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