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castortr0y [4]
3 years ago
8

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/m

ol. How much heat is required to vaporize 250 g of ethanol at 78.5 °C? Ans = 209.8 KJ
Chemistry
1 answer:
garik1379 [7]3 years ago
6 0

Answer:

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

Explanation:

Mass of an ethanol = 250 g

Molar mass of ethanol = 46 g/mol

Moles of an ethanol = n=\frac{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\times \Delta H_{vap}

=5.435 mol\times 38.6 kJ/mol=209.7826 kJ\approx 209.8kJ

Q = 209.8 kilo Joules

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