What is the volume of one molecules of water (density of `H_(2) O = 1 g cm^(-3)`)
b. What is the radius of the water molecule assuming it be spherical.
c. Calculate the radius of the oxygen atom, assuming the oxygen atom occupies half of the volume occupied by the water molecule.

Explanation

a. mole of `H_(2)O = 18 g = 18 cm^(3)`
(`:' d_(H_(2)^(2)) = 1 g cm^(-3))`
`= 6.022 xx 10^(23)` molecules of `H_(2) O`
There,fore, volume of 1 molecules of `H_(2) O`
`= (18)/(6.022 xx 10^(23)) cm^(3)`
`= 2.989 xx 10^(-23) cm^(3)`
b. Since `H_(2) O` molecules is spherical, if its radius is r, then volume will be
`(4)/(3) pi r^(3) = 2.989 xx 10^(-23) cm^(3)`
or `r^(3) = 7.133 xx 10^(-24)`
or `r = (7.133 xx 10^(-24))^(1)/(3) = (7.133)^(1)/(3) xx 10^(-8) cm`
Put `x = (7 .133)^(1)/(3)`
`log x = (1)/(3) log. 7.133 = (1)/(3) xx 0.8533 = 0.2844`
`x =` Antilog `0.2844 = 1.925`
`:. r = 1.925 xx 10^(8) cm`
c. Let `R` is the radius of the oxygen atom, and the oxygen atom occupies half the volume of `H_(2)O` molecule, then
`(4)/(3) pi R^(3) = (1)/(2) xx 2.989 xx 10^(-23) cm^(3)`
`R^(3) = 3.566 xx 10^(-24) cm^(3)`
`R = 1.528 xx 10^(-8) cm`