Question

The ratio of volume occupied by 1 mole water vapour at NTP & 1 MOL liquid water is: With solution ​

Answer 1

Answer:

22400 : 18

Explanation:

Volume occupied by 1 mol water vapour= 22400ml

The density of liquid water is 1 g/ml

volume= mass/ density

therefore, Volume of 1 mole of water vapour

= (18g)/(1 g/ml)= 18ml

So, The ratio of volume occupied by 1 mol water vapour at NTP and 1 mol liquid water is :-

22400 : 18

Answer 2

The ratio is 11200:9.

Given - Number of moles

Find - Ratio of the volume occupied by 1 mole water vapour at NTP & 1 MOL liquid water.

Solution - As a known fact, the volume occupied by 1 mole water vapour at NTP is 22400 ml. Also, as per the known fact, the volume of 1 mole of water is 18 ml. So, their ratio is 22400:18.

Converting it in fraction form -

Volume = $\frac{22400}{18}$

Performing division

Volume = $\frac{11200}{9}$

Hence, the ratio of volume is 11200:9.

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