One mole of NaCl (s) on melting absorbed 30.5 kJ one of heat and its entropy is increased by 28.8 JK-1mol-1. The melting point of NaCl is around

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1059 K

30.5 K

28.8 K

28800 K

answer is A.

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Detailed Solution

∆S=28.8JK-1mol-1=28.8×10-3kJK-1mol-1NaCl(s) →NaCl(1) Given ∆H = 30.5 kJ mol-1By using ∆S=∆HT=30.528.8×10-3=1059K

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