Enthalpy of fusion of a liquid is 1.435kcalmol−1 and molar entropy change is 5.26calmol−1K−1, hence melting point of liquid is
A
100∘C
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0∘C
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
373K
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
−273∘C
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B0∘C Given: △Hfusion=1.435kcalmol−1 =1435calmol−1 △S=5.26calmol−1k−1 we know: △Sfusion=△HfusionTT=△Hfus△Sfus=14355.26=272.81K=(272.81−273)oC≈0∘C