Question

When 8.70 kJ of thermal energy is added to 2.50 mol of liquid methanol, it vaporizes. Determine the heat of vaporization in kJ/mol of methanol.

Answer 1

The heat of vaporization in kJ/mol of methanol

= 3.48 kJ/mole

Step-by-step explanation:

Here,the heat of vaporization in kJ/mol is asked to calculate.This means it is asked to calculate heat required for 1 mole of methanol. Since the substance vaporizes , so this is called heat of vaporization.

Divide the thermal energy with the number of moles .

This is simple mathematics :

If 2.50 mol of liquid require = 8.70 kJ of energy

Then, 1 mole will need =

`(8.70)/(2.50)`

= 3.48 kJ/mole

Follow the units :

The heat of vaporization in kJ/mol of methanol is asked.

`Heat\ of\ Vaporization=(Thermal\ energy)/(Moles)`

`=(kJ)/(mole)`

Unit = kJ /Mole