Question

 (1) 18 mL water
As dH₂O = 1 g/mL So WH₂O = 18g
$\eta$H₂O =$\frac{18}{18}=1$
molecules = 1 × NA
(2) 0.18 g of water
$\eta$H₂O =$\frac{0.18}{0.18}=0.01$
(molecules)H₂O = 0.01 × NA
(3) (VH₂O(g))STP = 0.00224 L
$\eta$H₂O=$\frac{V}{22.4}=\frac{0.00224}{22.4}=0.0001$
molecules = 0.0001 × NA
(4) nH₂O = 10^–3
${\left(molecules\right)}_{H_{2}O}$ = 10^–3 × NA