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Q. Number of molecules in one litre of water is close to:

J & K CETJ & K CET 2000

Solution:

$1 \,L\,H _{2} O =1000\, cc =1000\, g$
$\therefore 18\, g$ of $H_{2} O =6.023 \times 10^{23}$ molecules
$\therefore 1000\, g$ of $H_{2} O =\frac{6.023 \times 10^{23} \times 1000}{18}$
$=6.023 \times 10^{23} \times 55.5$ molecules