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).

Explanation

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`