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 = 
= 3.48 KJ
So, the heat of vaporization 
Therefore, the heat of vaporization of methanol is 3.48KJ/Mol
Thus we can balance the oxygen atoms by putting a prefix of 25/2 on the left side. To obtain a equation containing whole numbers, we multiply the entire equation by 2. This gives the final equation. 2 C8H18 + 25 O2 ---> 16 CO2 +18 H2O.
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