Find the number of atoms of hydrogen and oxygen present in one spherical drop of water with radius 1 mm at `4^(@)C`.

Explanation

The volume of one spherical drop of water
`=(4)/(3) xxpixx(0.1)^(3)=(4)/(3)xx(22)/(7)xx10^(-3)=4.19xx10^(-3)cm^(3)`
Further, density of water at `4^(@)C=1g" "cm^(-3)`
`therefore`Mass of `4.19xx10^(-3)cm^(3)` water at `4^(@)C=4.19xx10^(-3)g`.
Now, mass of 1 gram-molecule of water=18g.
`therefore`Number of molecules in 18g water=`6.022xx10^(23)`
`therefore`Number of molecules present in `4.19xx10^(-3)`g o water
`=(6.022xx10^(23)xx4.19xx10^(-3))/(18)=1.417xx10^(20)`
1 molecule of water contains 2 hydrogen and 1 oxygen atom.
`therefore`The number of hydrogen atoms in `1.4017xx10^(20)`
molecules of water=`2xx1.4017xx10^(20)=2.803xx10^(20)`
so, the number of oxygenn atoms present in `1.4017xx10^(20)`
molecules of water=`1.4017xx10^(20)`.