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Q.

A substance A has a normal melting point of 250K and normal boiling point of 300 K. Using this information and other information given below ΔSsublimation  at 250K [in cal/K-mole] 

 [Given : R=2cal/Kmole]

Information-1: Entropy of vapourisation at 300 K is 21cal/Kmole

 Information-2: Latent heat of fusion at 250 K is 2.5kcal/mole

 Information-3: Cp is liquid A and gaseous A is 20  cal/K mole  and 10cal/Kmole

 Information-4: In 63=0.18 

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answer is 33.

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Detailed Solution

A(s)A(g);ΔS300K=10.5R

ΔS300=ΔS250K+ΔrxnnCp,mln30025010.5R=ΔS250K+(10)ln65

10.5R=ΔS250K0.18×1021+1.8=ΔS250KΔS250K=22.8cal/moleKΔSTotal =22.8+2500250=32.833

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