The entropy change for the melting of ‘x’ moles of ice (heat of fusion is 80 cal g–1) at 273 K and 1 atm pressure is 28.80 cal K–1

Q. The entropy change for the melting of ‘x’ moles of ice (heat of fusion is 80 cal g^–1) at 273 K and 1 atm pressure is 28.80 cal K^–1. The value of ‘x’ is___________. (Round off to two decimal places)

(Molecular weight of water =18 g/mol)

Solution:

Δ S = Δ H/ T , OR 

Δ S = Δ H x W / T , where W = weight. 

No. of Moles = Given Mass / Molar mass

Calculation:- 

Δ S = Δ H x W / T,

28.80 cal K-1 = 80 cal g-1 x W / 273 K,

W = 28.80 cal K-1 x 273 K / 80 cal g-1 = 98.28 g. 

Using, No. of Moles = Given Mass / Molar mass ,

No of moles = 98.28 g / 18 g = 5.46 moles.