Calculate
(a) the actual volume of a molecule of water
(b) the radius of a water molecule assuming to be spherical (density of water `= 1 g cm"^(-3)`)

Explanation

(a) Density of water = `1g mL^(-1)`
`therefore` 1g of `H_(2)O = 1mL`
1 mole of water = 18 g of `H_(2)O` = 18 mL of water `6.022 xx 10^23` molecules of water occupy volume = 18 mL
1 molecules of water occupy volume `=18/(6.022 xx 10^(23))`
`=2.99 xx 10^(-23)` mL
(b) Since water molecule is assumed to be spherical, its
volume is equal to `4/3 pi r^(3)`, where r is the radius of water molecules.
`therefore 4/3 pir^(3) = 2.99 xx 10^(-23)`
=1.925 `Å`