Question

Ethanol, CH.0, is common beverage alcohol. At its boiling point of 78.5 °C, the enthalpy of vaporization of ethanol is 38.6 kJ/mol. How much heat is required to vaporize 250 g of ethanol at 78.5 °C? Ans = 209.8 KJ

Answer 1

209.8 kilo Joules heat is required to vaporize 250 g of ethanol at 78.5 °C.

Step-by-step explanation:

Mass of an ethanol = 250 g

Molar mass of ethanol = 46 g/mol

Moles of an ethanol =
`n=(250 g)/(46 g/mol)=5.435 mol`

Enthalpy of vaporization of ethanol =
`\Delta H_(vap) =38.6 kJ/mol`

Heat required to vaporize 250 g of ethanol at 78.5 °C : Q

`Q=n* \Delta H_(vap)`

`=5.435 mol* 38.6 kJ/mol=209.7826 kJ\approx 209.8kJ`

Q = 209.8 kilo Joules