Question

Convert 1 mole of liquid water (5 atm 373 k) into vapour (5 atm 373 k)

Answer 1

1 mole water = W / 18 amu

W = 18g = 18ml.=V(vol.) {as density of water = 1g/cm³ and 1cm³ = 1ml }

1 mole vapour ; using IDEAL gas equation

PV = nRT

5 × V = 1 × 0.0821 × 373

V = 30.62/5

V = 6.124 litre => 6124 ml

Thus 1 mole of liquid water occupies 18ml

and 1 mole of water vapour ( at 5 atm , 373K ) occupies 6124 ml

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