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 cm^(-3)`
Gram molecular mass of water `= 18 g`
Gram molecular volume `= ((18g))/((1g cm^(-3)))=18 cm^(3)`
`6.022 xx 10^(23)` molecules of water occupy volume `= 18 cm^(3) = 18 c c`
1 molecule of water occupies volume `= (18)/(6.022xx10^(23))=2.99xx10^(-23) c c`
(b) Volume of a sphere `= 4//3 pi r^(3) :. 4//3 pi r^(3) = 2.99 xx 10^(-23) c c`
or `r^(3)=(2.99xx10^(-23)xx3)/(4xx3.143)" "[becausepi=22//7=3.143]`
or `r=((2.99xx10^(-23)xx3)/(4xx3.143))^(1//3)=(7.13xx10^(-24))^(1//3)=1.925xx10^(-8)"cm"=1.925 Ã…`.