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

Answer:

The heat of vaporization in kJ/mol of methanol

= 3.48 kJ/mole

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 =

$\frac{8.70}{2.50}$

= 3.48 kJ/mole

Follow the units :

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

$Heat\ of\ Vaporization=\frac{Thermal\ energy}{Moles}$

$=\frac{kJ}{mole}$

Unit = kJ /Mole