(A) Volume of one drop $= (\frac{1}{25}) \ mL$$\therefore$ Mass of $1$ drop $= V \times d = (\frac{1}{25} \ mL)(1 \ g / mL) = \frac{1}{25} \ g$Number

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(A) Volume of one drop $= (\frac{1}{25}) \ mL$$\therefore$ Mass of $1$ drop $= V \times d = (\frac{1}{25} \ mL)(1 \ g / mL) = \frac{1}{25} \ g$Number of moles of $H_2O = \frac{\text{Mass of water in one drop}}{\text{Molar mass of water}} = \frac{1/25}{18} = \frac{1}{450} \ mol$$\therefore$ Number of $H_2O$ molecules $= \frac{1}{450} \ N_0 = \frac{0.02}{9} \ N_0$

Class 11 Chemistry Some Basic Concepts of Chemistry The mole concept

Easy

$1 \ mL$ of water has $25$ drops. Let $N_0$ be the Avogadro number. What is the number of molecules present in $1$ drop of water? (Density of water $= 1 \ g / mL$)

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A
$\frac{0.02}{9} \ N_0$
B
$\frac{18}{25} \ N_0$
C
$\frac{25}{18} \ N_0$
D
$\frac{0.04}{25} \ N_0$

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Some Basic Concepts of Chemistry

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The mole concept

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$1 \ mL$ of water has $25$ drops. Let $N_0$ be the Avogadro number. What is the number of molecules present in $1$ drop of water? (Density of water $= 1 \ g / mL$)

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