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.
Please answer quickly, its urgent!

Answer 1

Answer:

The heat of vaporisation of methanol is "3.48 KJ/Mol"

Explanation:

The amount of heat energy required to convert or transform  1 gram of liquid to vapour is called heat of vaporisation

When 8.7 KJ of heat energy is required to vaporize 2.5 mol of liquid methanol.

Hence, for 1 mol of liquid methanol, amount of heat energy required to evaporate the methanol is =   $\frac{8.7}{2.5}KJ$

 = 3.48 KJ

So, the heat of vaporization $\delta H_{vap} = 3.48KJ/Mol$

Therefore, the heat of vaporization of methanol is 3.48KJ/Mol