Question

At 100 C and 1 atm, if the density of liquid water is 1.0 g cm−3 and that of water vapour is 0.0006 g cm−3, then the volume occupied by water molecules in one litre of steam at that temperature is :

Answer 1

Answer:

0.60cm3

Explanation:

Let us consider, 1.0 L of liquid water is converted into steam volume of H2O (l) = 1 L, mass = 1000 g

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

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

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

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 in1000 cm3 steam = (1000 )/1000 × 0.0006 × 1000

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 in1000 cm3 steam = (1000 )/1000 × 0.0006 × 1000= 0.60 cm3

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