The number of water molecules present in a dr | vedclass

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Question

(A) Density of water at room temperature is $1 \ g/mL$. Mass of the drop $= 	ext{Density} 	imes 	ext{Volume} = 1 \ g/mL 	imes 0.0018 \ mL = 0.0018

Vedclass · Accepted Answer

$6.023 	imes 10^{19}$

Vedclass · Answer

(A) Density of water at room temperature is $1 \ g/mL$. Mass of the drop $= 	ext{Density} 	imes 	ext{Volume} = 1 \ g/mL 	imes 0.0018 \ mL = 0.0018 \ g$. Molar mass of water $(H_2O)$ $= 18 \ g/mol$. Number of moles $= \frac{	ext{Mass}}{	ext{Molar mass}} = \frac{0.0018 \ g}{18 \ g/mol} = 1 	imes 10^{-4} \ mol$. Number of water molecules $= 	ext{Number of moles} 	imes N_A = 1 	imes 10^{-4} 	imes 6.023 	imes 10^{23} = 6.023 	imes 10^{19}$.

The number of water molecules present in a drop of water (volume $0.0018 \ mL$) at room temperature is

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