The number of atoms present in a drop of wate | vedclass

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(D) The molar mass of water $(H_2O) = (2 	imes 1) + 16 = 18 \ g/mol$. The number of moles in $1.8 \ g$ of water $= \frac{1.8 \ g}{18 \ g/mol} = 0.1 \

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$3 	imes \frac{N_A}{10}$

Vedclass · Answer

(D) The molar mass of water $(H_2O) = (2 	imes 1) + 16 = 18 \ g/mol$. The number of moles in $1.8 \ g$ of water $= \frac{1.8 \ g}{18 \ g/mol} = 0.1 \ mol = \frac{1}{10} \ mol$. The number of molecules of water $= 	ext{moles} 	imes N_A = \frac{N_A}{10}$. Each molecule of water $(H_2O)$ contains $3$ atoms ($2$ hydrogen atoms and $1$ oxygen atom). Therefore,the total number of atoms $= 3 	imes 	ext{number of molecules} = 3 	imes \frac{N_A}{10}$.

The number of atoms present in a drop of water weighing $1.8 \ g$ is.....

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