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At 100^(@)C and 1 atm, if the density ...

At `100^(@)C` and 1 atm, if the density of the liquid water is `1.0 g cm^(-3)` and that of water vapour is `0.00006g cm^(-3)`, then the volume occupied by water molecule in 1 L steam at this temperature is:

A

`6 cm^(3)`

B

`60 cm^(3)`

C

`0.6 cm^(3)`

D

`0.06 cm^(3)`

Text Solution

Verified by Experts

The correct Answer is:
C
`1 L = 1000 mL = 1000 cm^(3)` Mass = Density `xx` volume
`= (0.006 g cm^(-3)) xx (1000 cm^(3))`
`= 0.6 g`
18 g of water `= 18 cm^(3)`
`:.` 0.6 g of water `= 0.6 cm^(3)`
`:.` 0.6 g of water `= 0.6 cm^(3)`
Actual volume occupied by molecules `= 0.6 cm^(3)`
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