Question

48. Latent heat of fusion of ice is 0.333 kJ g.?. The increase in entropy when 1 mole water melts at 0°C will be
(1) 21.98 kJ K mol-1
(2) 21.98 kcal K-mol-
(3) 21.98 J K mol
(4) 21.98 cal K-mol-1​

Answer 1

Answer:

option 3,

21.98 J/K mol

Explanation:

Method to Solve :

Latent heat of fusion of ice = 0.333 k J/g = 333 J/g

Amount of ice = 1 mole

Mass of 1 mole of ice will be the same as mass of 1 mole of water.

So,

Mass of 1 mole of ice = 2 (for 2 moles of H₂ atoms) + 16 (for 1 mole of O atom)

Mass of 1 mole of ice = 18 g

Temperature = T = 0�C = 273.15 K

Change in entropy = ?

Solution:

Entropy change = ΔS = n ΔH / T

Entropy change = ΔS = (18) ( 333) / 273.15

Entropy change = ΔS = 21.98 J/K mol