The entropy change for the conversion of the one gram of ice to water at 273 K and one atmospheric pressure - H fusion -6-025 kJ mol -1- is-A- 334-72 J g-1B- 1-226 k J K-1 g-1C- 913-73 k J K-1 g-1D- None

The correct option is B Δ H_fusion=6.025kJ mol^ 1=6025 J mol^ 1 =(6025 mol^ 1)/(18g mol^ 1)=334.72 Jg^ 1 [ ∵ Molar mass of H_2O=18g mol^ 1 ] Δ H_fusion=Δ H_ ...

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Byju's Answer

Standard XII

Chemistry

Third Law of Thermodynamics

The entropy c...

Question

The entropy change for the conversion of the one gram of ice to water at 273 K and one atmospheric pressure $[Δ H_{f u s i o n} = 6.025 k J m o l^{- 1}]$ is:

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Solution

The correct option is B
$Δ H_{f u s i o n} = 6.025 k J m o l^{- 1} = 6025 J m o l^{- 1} = \frac{(6025 m o l^{- 1})}{(18 g m o l^{- 1})} = 334.72 J g^{- 1} [∵ M o l a r m a s s o f H_{2} O = 18 g m o l^{- 1}] Δ H_{f u s i o n} = \frac{Δ H_{f u s i o n}}{T_{f}} = \frac{(334.72 J g^{- 1})}{(273 K)} = 1.226 J K^{- 1} g^{- 1}$

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Similar questions

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The entropy change for the conversion of the one gram of ice to water at 273 K and one atmospheric pressure [ΔHfusion=6.025kJmol−1] is:

The correct option is B ΔHfusion=6.025kJmol−1=6025Jmol−1=(6025mol−1)(18gmol−1)=334.72Jg−1[∵Molar mass of H2O=18gmol−1]ΔHfusion=ΔHfusionTf=(334.72Jg−1)(273K)=1.226JK−1g−1

The entropy change for the conversion of the one gram of ice to water at 273 K and one atmospheric pressure $[Δ H_{f u s i o n} = 6.025 k J m o l^{- 1}]$ is: