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Molal enthalpy of fusion of water at 273...

Molal enthalpy of fusion of water at 273 is 6.0246 kJ `mol^(-1)`. Calculate mole depression constnt for water.

Text Solution

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`K_(f)=(MR(T_(f)^(@))^(2))/((DeltaH_(fusion)))`
`DeltaH_(fusion)=6.0246 kJ mol^(-1)=6024.6 J mol^(-1)`
`M=18 g mol^(-1)=0.018 kg mol^(-1)`
`R=8.314 JK^(-1)mol^(-1)=6024.6Jm0l^(-1)`
M=18 g mol^(-1)=0.018 kg mol^(-1)
`R=8.314 JK^(-1)mol^(-1),T_(f)^(@)=273 K`
`K_(f)=((0.018 kg mol^(-1))xx(8.314 JK^(-1)mol^(-1))xx(273K)^(2))/((6024.6J mol^(-1)))`
=1.85 K kg mol^(-1)=1.85 Km^(-1)`
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