The normal boiling point of water is 373 Kelvin.vapour pressure of water at 298 kelvin is 23 mm .if enthalpy of vaporization is 40.656 kg per mole the boiling point of water at 23 mm pressure will be

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Arun · Answer

From the Calsius Clapeyron formulae: P = A exp (- D H vap / R T ) P = 23exp(-40656/8.3145*(298 – 373)) = 23exp(65.19) K Regards Arun (askIITians forum expert)

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The normal boiling point of water is 373 Kelvin.vapour pressure of water at 298 kelvin is 23 mm .if enthalpy of vaporization is 40.656 kg per mole the boiling point of water at 23 mm pressure will be

The normal boiling point of water is 373 Kelvin.vapour pressure of water at 298 kelvin is 23 mm .if enthalpy of vaporization is 40.656 kg per mole the boiling point of water at 23 mm pressure will be

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The normal boiling point of water is 373 Kelvin.vapour pressure of water at 298 kelvin is 23 mm .if enthalpy of vaporization is 40.656 kg per mole the boiling point of water at 23 mm pressure will be

The normal boiling point of water is 373 Kelvin.vapour pressure of water at 298 kelvin is 23 mm .if enthalpy of vaporization is 40.656 kg per mole the boiling point of water at 23 mm pressure will be

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