The number of molecules present in 3.6 g of water (H2O, M = 18 g mol^-1) is (use N_A = 6.02x10^23 mol^-1):

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The number of molecules present in $3.6 g$ of water ( $H X_{2} O$ , $M = 18 g mo l^{- 1}$ ) is (use $N_{A} = 6.02 \times 1 0^{23} mo l^{- 1}$ ):

Moles of water

= 3.6/18 = 0.20

. Number of molecules

= 0.20 \times 6.02 \times 1 0^{23} = 1.204 \times 1 0^{23}

. The value with a factor 3 counts atoms, not molecules.