(a) 18 mL water :
Since, density of H₂O = 1 g/mL
Mass of water = V × d = 18 × 1 = 18 g
No. of molecules of H₂O = ${{18} \over {18}} \times$ N_A = 1 $\times$ N_A

(b) 0.18 g of water :
No. of molecules of H₂O = ${{0.18} \over {18}} \times$ N_A = 0.01$\times$ N_A

(c) Volume of H₂O(g) STP = 0.00224 L
No. of molecules of H₂O = ${V \over {22.4}}$ $\times$ N_A
= ${{0.00224} \over {22.4}}$ $\times$ N_A = 0.0001 $\times$ N_A

(d) No of Molecules of water = mole × N_A = 10^–3 N_A