At 100°C and 1 atm, if the density of liquid water is 1.0 g cm^(-3) andthat of water vapour is 0.0006 gcm-3 , then the volume occupied by water molecules in 1 litre of steam at that temperature is:

At 100°C and 1 atm, if the density of liquid water is 1.0 g cm^(-3) a

1.	At 100°C and 1 atm, if the density of liquid water is 1.0 g cm^(-3) andthat of water vapour is 0.0006 gcm-3 , then the volume occupied by water molecules in 1 litre of steam at that temperature is:
Answer» `6 cm^(3)` `60 cm^(3)` `0.6 cm^(3)` `0.06 cm^(3)` Solution :Mass of 1 LITRE of water = 1000 g (`therefore d = 1 g cm^(3)` ) The volume of steam FORMED by 1000 g of water `V=(nRT)/P = 100/18 xx (0.0821 xx 373)/1 = 1701.3 L` Thus, 1701.3 L of steam has a mass of 1000 g. `therefore` Mass of 1 L of steam `=1000/(1701.3) = 0.6 g` Thus, 1L of steam contains only 0.6 g of water. Since, density of water under given conditions is `1 g cm^(-3)`, the volume occupied by water molecules in 1 L of steam would be `0.6 cm^(-3)`.