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Thermodynamics question

2012 · Q119
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Thermodynamics question

2012 · Q119

NEETChemistryThermodynamicsMCQ+4 / −1
The enthalpy of fusion of water is 1.435 kcal/mol. The molar entropy change for the melting of ice at 0oC is
  1. A
    10.52 cal/(mol K)
  2. B
    21.04 cal/(mol K)
  3. C
    5.260 cal/(mol K)
  4. D
    0.526 cal/(mol K)
View written solutionFree

Correct answer: C

H2O(lll) → H2O(s)

∆H = 1.435 Kcal/mol

T = 0 + 273K = 273K

ΔS=ΔHT\Delta S = {{\Delta H} \over T}ΔS=TΔH​

⇒\Rightarrow⇒ ΔS=1.435273\Delta S = {{1.435} \over {273}}ΔS=2731.435​ = 5.26 ×\times× 10-3 kcal/mol K

⇒\Rightarrow⇒ ΔS\Delta SΔS = 5.260 cal/mol K

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