At 1000C and 1 atm, if the density of liquid water is 1.0 g/cc and that of water vapour is 0.0006 g/cc, then the volume occupied by water molecule in 1 litre of steam at that temperature is

# At ${100}^{0}\mathrm{C}$ and 1 atm, if the density of liquid water is 1.0 g/cc and that of water vapour is 0.0006 g/cc, then the volume occupied by water molecule in 1 litre of steam at that temperature is

1. A

6 cc

2. B

60 cc

3. C

0.6 cc

4. D

0.06 cc

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### Solution:

Mass of 1 L water vapour = V x d = 1000 x 0.0006 = 0.6 g

$\therefore$  volume of liquid water =

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