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Volume occupied by one molecule of water...

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

A

`3.0 xx 10^(-23) cm^(3)`

B

`5.5 xx 10^(23) cm^(3)`

C

`9.0 xx 10^(-23) cm^(3)`

D

`6.023 xx 10^(-23) cm^(3)`

Text Solution

Verified by Experts

The correct Answer is:
A
Mass of one molecule of water
`= ((18g))/(6.022xx10^(23))=3xx10^(-23)g`
Since density of water is 1g
`:.` Volume occupied by one molecule of water
`= 3 xx 10^(-23)` cc.
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What is the volume of one molecules of water (density of H_(2) O = 1 g cm^(-3) ) b. What is the radius of the water molecule assuming it be spherical. c. Calculate the radius of the oxygen atom, assuming the oxygen atom occupies half of the volume occupied by the water molecule.

Knowledge Check

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

    A
    `3xx10^(-23)cm^(3)`
    B
    `5.5xx10^(-23)cm^(3)`
    C
    `9xx10^(-23)cm^(3)`
    D
    `6.023xx10^(-23)cm^(3)`

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

    A
    `3.0 xx 10^(-23) cm^(3)`
    B
    `5.5 xx 10^(-23) cm^(3)`
    C
    `9.0 xx 10^(-23) cm^(3)`
    D
    `6.023 xx 10^(-23) cm^(3)`

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

    A
    (a)`18 cm^(3)`
    B
    (b)`22400 cm^(3)`
    C
    (c )`6.02xx10^(-23) cm^(3)`
    D
    (d)`3.0xx10^(-23) cm^(3)`

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    (i) Assuming the density of water to be 1g//cm^(3) , calculate the volume occupied by one molecule of water. (ii) Assuming the water molecule to be spherical, calculate the diameter of the water molecule. (iii) Assuming the oxygen atom occupied half of the volume occupied by the water molecule, calculate approximately the diameter of the oxygen atom.

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