Question

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?

Answer 1

`100g` benzene contains `=(100)/(78) mol e`
`( mol. Wt. ` of benzene,` C_(6)H_(6)=78)`
`:. `Heat of vaporisation of `(100)/(78)` mole ` C_(6)H_(6)`
`=(100)/(78)xx30.8kJ=(100xx10^(3)xx30.8)/(78)J`
Since, Power `=(en ergy)/(time)(1W=1J//sec)`
`:. 100=(100xx10^(3)xx30.8)/(78xxtime)`
`:. time =394.87 sec=(394.87)/(60)mi n`
`=6.6 mi n`