Standard vaporization enthalpy of benzene at its boiling point is `30.8 kJ mol^(-1)`, for how long would a `100W` electric heater have to operate in order to vaporize a `100g` sample of benzene at its boiling temperature?

The molar enthalpy of vaporization of benzene at its boiling point (353 K) is 29.7 KJ//"mole" . For how long minute) would a 11.4 Volt source need to supply a 0. 5A current in order to vaporise 7.8 g of the sample at its boiling point ?

The enthalpy of vaporization of a certain liquid at its boiling point of 35^(@)C is 24.64 kJ "mol"^(-1) .The value of change in entropy for the process is

Standard enthalpy of vaporization of water at 1 atm pressure and 100°C is 40.66 kJ mol^(-1) . The internal energy change of vaporization of 2 moles of water at 100°C (in kJ) is

It the enthalpy of vaporization of benzene is 308/ kJ mol^(-1) at boiling point (80^(@)C) , calculate the entroy (j mol^(-1) K^(-1)) in changing it from liquid to vapor.

Enthalpy of vapourization of benzene is +35.3 kJ mol^(-1) at its boiling point of 80^(@)C . The entropy change in the transition of the vapour to liquid at its boiling points [in K^(-1) mol^(-1) ] is

The enthalpy of vaporization of a substance is 840 J per mol and its boiling point is -173^@C . Calculate its entropy of vaporization.

The normal boiling point of toluene is 110.7^(@)C and its boiling point elevation constant 3.32 " K kg mol"^(-1) . The enthalpy of vaporization of toluene is nearly :