If the density of water is 1 , g , cm^{-3},th | vedclass

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(D) The molar mass of water $(H_2O)$ is $18 \, g \, mol^{-1}$.Given density $(d)$ = $1 \, g \, cm^{-3}$.Since $d = \frac{mass}{volume}$,the volume of

Vedclass · Accepted Answer

$3.0 	imes 10^{-23} \, cm^3$

Vedclass · Answer

(D) The molar mass of water $(H_2O)$ is $18 \, g \, mol^{-1}$.Given density $(d)$ = $1 \, g \, cm^{-3}$.Since $d = \frac{mass}{volume}$,the volume of $1 \, mole$ $(18 \, g)$ of water is $V = \frac{18 \, g}{1 \, g \, cm^{-3}} = 18 \, cm^3$.$1 \, mole$ of water contains $6.022 	imes 10^{23}$ molecules.Therefore,the volume occupied by $1$ molecule is $\frac{18 \, cm^3}{6.022 	imes 10^{23}} \approx 2.99 	imes 10^{-23} \, cm^3 \approx 3.0 	imes 10^{-23} \, cm^3$.

If the density of water is $1 \, g \, cm^{-3}$,then the volume occupied by one molecule of water is approximately:

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