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

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

Step-by-step 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 =
`(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