If the density of water is 1 g `cm^(-3)` then the volume occupied by one molecule of water is approximately

Options

A. `18 cm^(3)`
B. `22400 cm^(3)`
C. `6*02xx10^(-23) cm^(3)`
D. `3*0xx10^(-23) cm^(3)`

Answer

C

Explanation

Mass of 1 molecule of water
`=("GMM")/(N_(0))=(18)/(6*02xx10^(23))g`
`:.` Volume of 1 molecule of water `= ("Mass")/("Density")`
`=(18xx10^(-23))/(6*02xx1)=3*0xx10^(-23) mL`.