Calculate the entropy change during the melting of one mole of ice into water at `0^@C` and 1 atm pressure. Enthalpy of fusion of ice is 6008 J `"Mole"^(-1)`

Options

A. `22.007 "J K"^(-1) "Mole"^(-1)`
B. 22.007 J K Mole
C. `220.07 "J K"^(-1) "Mole"^(-1)`
D. `2.2007 "J K"^(-1)` Mole

Answer

A

Explanation

Enthalpy of fusion of ice = 6008 J `"mol"^(-1) =DeltaH`
`DeltaS=(DeltaH)/T_m " " T_m =0^@C` = 273 K
Entropy of fusion = `"Enthalpy of fusion of ice "/"Melting point (Kelvin)"`
`DeltaS=6008/273 =22.007 J K^(-1) "mol"^(-1)`