Volume occupied by one molecule of water (density = 1 g `cm^(-3)`)A. `3.0xx10^(-23)cm^(3)`B. `5.5xx10^(-23)cm^(3)`C. `9.0xx10^(-23)cm^(3)`D. `6.023xx10^(-23)cm^(3)`

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

1.	Volume occupied by one molecule of water (density = 1 g `cm^(-3)`)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» Correct Answer - A 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)`

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Volume occupied by one molecule of water (density = 1 g `cm^(-3)`)A. `3.0xx10^(-23)cm^(3)`B. `5.5xx10^(-23)cm^(3)`C. `9.0xx10^(-23)cm^(3)`D. `6.023xx10^(-23)cm^(3)`

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