1.

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

Answer»

`9.0xx10^(-23)cm^(3)`
`6.023xx10^(-23)cm^(3)`
`3.xx10^(-23)cm^(3)`
`5.5xx10^(-23)cm^(3)`

Solution :`"1 mole of "H_(2)O=6.02xx10^(23)" molecules"`
`=18 g = 18 cm^(3)`
`therefore" Volume OCCUPIED by 1 molecule of "H_(2)O`
`=(18)/(6.02xx10^(23))cm^(3)=3xx10^(-23)cm^(3)`


Discussion

No Comment Found

Related InterviewSolutions