The volume of water molecule is
Take, density of wter is `10^(3) kg m^(-3)` and avogadros' number `=610^(23) "mole"^(-1))`

Options

A. `3xx10^(-28)m^(3)`
B. `3xx10^(-29)m^(3)`
C. `1.5xx10^(-28)m^(3)`
D. `1.5xx10^(-29)m^(3)`

Answer

B

Explanation

Molecular mass of water = 28
`therefore` Number of molecules in 18 g
or 0.018 kg of water `=6xx10^(23)`
`therefore` Mass of molecule of water `=(0.018)/(6xx10^(23))=3xx10^(-26)`
`therefore` Volume of a water molecule
`=("Mass")/("Density")=(3xx10^(-26))/(1000)=3xx10^(-29)m^(3)`