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Question

(D) At $STP$,the molar volume of an ideal gas is $22400 \ mL \ mol^{-1}$.Number of moles of water vapour in $1 \ mL$ is $n = \frac{1 \ mL}{22400 \ mL

Vedclass · Accepted Answer

$2.69 	imes 10^{19}$

Vedclass · Answer

(D) At $STP$,the molar volume of an ideal gas is $22400 \ mL \ mol^{-1}$.Number of moles of water vapour in $1 \ mL$ is $n = \frac{1 \ mL}{22400 \ mL \ mol^{-1}} = 4.464 	imes 10^{-5} \ mol$.The number of molecules is given by $N = n 	imes N_A$,where $N_A = 6.022 	imes 10^{23} \ mol^{-1}$.$N = 4.464 	imes 10^{-5} 	imes 6.022 	imes 10^{23} \approx 2.69 	imes 10^{19}$ molecules.Thus,the correct option is $D$.

Find the number of water molecules in $1 \ mL$ of water vapours at $STP$?

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