Question

The latent heat for melting of ice is 80 cal/g at 0ºC and 1 atm. Calculate its value in S.I. unit.​

Answer 1

We know,

We know,K

We know,K f

We know,K f

We know,K f (H

We know,K f (H 2

We know,K f (H 2

We know,K f (H 2 O)=

We know,K f (H 2 O)= ΔH

We know,K f (H 2 O)= ΔH fusion

We know,K f (H 2 O)= ΔH fusion

We know,K f (H 2 O)= ΔH fusion m×1000

We know,K f (H 2 O)= ΔH fusion m×1000RT

We know,K f (H 2 O)= ΔH fusion m×1000RT f

We know,K f (H 2 O)= ΔH fusion m×1000RT f

We know,K f (H 2 O)= ΔH fusion m×1000RT f

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent =

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×1000

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×10002×(273.15)

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×10002×(273.15) 2

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×10002×(273.15) 2 ×18

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×10002×(273.15) 2 ×18

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×10002×(273.15) 2 ×18

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×10002×(273.15) 2 ×18 =

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×10002×(273.15) 2 ×18 = 80×1000

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×10002×(273.15) 2 ×18 = 80×10002×74610.9

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×10002×(273.15) 2 ×18 = 80×10002×74610.9

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×10002×(273.15) 2 ×18 = 80×10002×74610.9

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×10002×(273.15) 2 ×18 = 80×10002×74610.9 =1.865K.Kg/mol

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×10002×(273.15) 2 ×18 = 80×10002×74610.9 =1.865K.Kg/molT

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×10002×(273.15) 2 ×18 = 80×10002×74610.9 =1.865K.Kg/molT f

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×10002×(273.15) 2 ×18 = 80×10002×74610.9 =1.865K.Kg/molT f′

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×10002×(273.15) 2 ×18 = 80×10002×74610.9 =1.865K.Kg/molT f′

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×10002×(273.15) 2 ×18 = 80×10002×74610.9 =1.865K.Kg/molT f′ =0

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×10002×(273.15) 2 ×18 = 80×10002×74610.9 =1.865K.Kg/molT f′ =0 o

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×10002×(273.15) 2 ×18 = 80×10002×74610.9 =1.865K.Kg/molT f′ =0 o C−iK

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×10002×(273.15) 2 ×18 = 80×10002×74610.9 =1.865K.Kg/molT f′ =0 o C−iK f

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×10002×(273.15) 2 ×18 = 80×10002×74610.9 =1.865K.Kg/molT f′ =0 o C−iK f

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×10002×(273.15) 2 ×18 = 80×10002×74610.9 =1.865K.Kg/molT f′ =0 o C−iK f .m

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×10002×(273.15) 2 ×18 = 80×10002×74610.9 =1.865K.Kg/molT f′ =0 o C−iK f .m =0−(1.95)(1.865)(

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×10002×(273.15) 2 ×18 = 80×10002×74610.9 =1.865K.Kg/molT f′ =0 o C−iK f .m =0−(1.95)(1.865)( 0.5

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×10002×(273.15) 2 ×18 = 80×10002×74610.9 =1.865K.Kg/molT f′ =0 o C−iK f .m =0−(1.95)(1.865)( 0.57.45/74.5

We know,K f (H 2 O)= ΔH fusion m×1000RT f 2 M solvent = 80×18×10002×(273.15) 2 ×18 = 80×10002×74610.9 =1.865K.Kg/molT f′ =0 o C−iK f .m =0−(1.95)(1.865)( 0.57.45/74.5

)

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