Find out number of heavy water molecules present in one drop of ordinary water whose volume is 0.05ml (density of H_2O = 1g/ml).

Find out number of heavy water molecules present in one drop of ordina

1.	Find out number of heavy water molecules present in one drop of ordinary water whose volume is 0.05ml (density of H_2O = 1g/ml).
Answer» SOLUTION :Weight of one DROP of water = density of water x VOLUME of water drop = 1 x 0.05 = 0.05 g 1g of `D_2O`is present per 6000 g of ordinary water. Weight of `D_2O` present in 0.05 g of ordinary water = `1/6000 xx 0.05 = 0.833 xx 10^(-5) g ` One GRAM mole of `D_2O` (20 g) contains ` 6.023 xx 10^23 D_2O` MOLECULES Number of `D_2O` molecules present in `0.833 xx 10^(-5) g. D_2O` which is present in a drop of water = `(0.833 xx 10^(-5))/(20) xx 6.023 xx 10^23 = 2.5 xx 10^17`