Determine the no. of molecules in a drop of water which weighs 0.09 g.

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Solution

Given mass of $H_{2} O$ = 0.09g
The molar mass of $H_{2} O$ = 18g
$M o l e = \frac{g i v e n m a s s}{m o l a r m a s s} = \frac{0.09}{18} = 0.005 m o l e$

$M o l e = \frac{n o . o f m o l e c u l e s}{A v o g a d r o n u m b e r}$

$N u m b e r o f m o l e c u l e s = N u m b e r o f m o l e s \times A v a g a d r o^{'} s n u m b e r$
$N u m b e r o f m o l e c u l e s = 0.005 \times 6.022 \times 10^{23}$
$N u m b e r o f m o l e c u l e s o f H_{2} O = 3. 011 \times 10^{21}$

Conclusive statement-

$N u m b e r o f m o l e c u l e s o f H_{2} O = 3. 011 \times 10^{21}$

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