Volume occupied by one molecule of water (density = 1 g `cm^(-3)`)

Options

A. `3.0xx10^(-23)cm^(3)`
B. `5.5xx10^(-23)cm^(3)`
C. `9.0xx10^(-23)cm^(3)`
D. `6.023xx10^(-23)cm^(3)`

Answer

A

Explanation

Weight of `6.023xx10^(23)` molecules of water = 18 g As volume occupied by `6.023xx10^(23)` molecules of water (dneisty `= "1 g cm"^(-3)`) will be
`=(18g)/(1gcm^(-3))`
`="18 cm"^(3)"or mL"`
So volume occupied by one molecule of water
`=(18)/(6.023xx10^(23))=2.988xx10^(-23)`
`=3.0xx10^(-23)cm^(3)`