![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](https://www.gkseries.com/blog/wp-content/uploads/2023/08/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.jpg)
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.