Question

at 100°C & 1 atm , if density of liquid water is 1 g cm⁻³ and that of water vapour is 0.0006 g cm⁻³ ,then volume occupied by water molecules in one litre of steam at that temperature is

Answer 1

Let us consider, 1.0 L of liquid water is converted into steam volume of H2O (l) = 1 L, mass = 1000 g
⇒ Volume of 1000 g steam = (1000 )/0.0006 cm3
∵ Volume of molecules in (1000 )/0.0006 cm3 steam = 1000 cm3
∴ Volume of molecules in
1000 cm3 steam = (1000 )/1000 × 0.0006 × 1000
= 0.60 cm3

Answer 2

Explanation:

60/100=6*10^-1 cc

volume of molecules is 1000/0.0006