1 mole of water $=6.022 \times 10^{23}$ molecules $=18$ gram of water
(1) $10^{-3} mol$ of water $=6.022 \times 10^{20}$ molecules
(2) $V =0.00224 L , P =1 atm , T =273 K$
$
\begin{aligned}
\therefore n =\frac{ PV }{ RT } &=\frac{1 \times 0.00224}{0.0821 \times 273}=10^{-4} \text { moles } \\
&=6.022 \times 10^{19} \text { molecules }
\end{aligned}
$
(3) $0.18 g$ of water $=10^{-2}$ moles
$
=6.022 \times 10^{21} \text { molecules }
$
(4) $18 ml$ of water $=18 g$ of water $ \because$ density of water $=1 \frac{ g }{ ml }$
$
\begin{array}{l}
=1 \text { mole } \\
=6.022 \times 10^{23} \text { molecules }
\end{array}
$
$\therefore 18 ml$ of water has the maximum number of water molecules.