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?

Explanation

Standard vaporisation enthalpy of benzene `=30.8 kJ mol^(-1)`
Power of electric heater `=100 W`
Molar mass of `C_(6)H_(6)=72+6=78 am u`
Heat required to vaporise `78 g C_(6)H_(6)=30.8xx1000J`
heat required to vaporise `100 g C_(6)H_(6)`
`=(30.8xx1000xx100)/78`
`=39900 J`
`100 W=100 J s^(-1)`
`100 J` heat is given in `1 s`
`39900` heat is given in `=(39900xx1)/(100)=399 s=6.6 min`