The total number of electrons present in 18 mL of water is how many time $N_A$ .\

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Professor PC Thomas Classes · Answer

$6.023 	imes 10^{24}$;\ $18 \mathrm{~mL} \mathrm{H}_{2} \mathrm{O}=18 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}=1 \mathrm{~mol}$ of $\mathrm{H}_{2} \mathrm{O}$\ 1 mol of $\mathrm{H}_{2} \mathrm{O}=10 	imes 6.023 	imes 10^{23}$\ $\left(\becauseight.$ No. of electrons in $\left.\mathrm{H}_{2} \mathrm{O}=2+8=10 e^{-}ight)$ \ $ =6.023 	imes 10^{24} 	ext { electrons } $