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 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 ?

Enthalpy of vaporisation of benzene is 30.8 kJ mol" at its normal boiling point of 80^(@)C. Time required for a 100 W electric heater to vapourise 100 g benzene at its boiling point (Given 1W = 1 Js^(-1) )

The enthalpy of vaporisation of benzene at 80^(@)C (boiling point) is 31 kJ*mol^(-1) . What will be the change in entropy of rthe transformation of 31.2g of benzene vapour into liquid benzene at 80^(@)C ?

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

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

The standard molar engthalpy of vaporisation of benzene Delta_(vap)H^(@) at 353 K is 30.8 kJ mol^(-1) . If the benzene vapours behave as an ideal gas, the change in internal energy of vaporisation of 78 g of benzene at 353 K in kJ mol^(-1) is