Question

How much energy is needed to vaporize 75.0 g of diethyl ether (c4h10o) at its boiling point (34.6°c), given that δhvap of diethyl ether = 26.5 kj/mol?

Answer 1

Answer: 26.8 kJ of energy is needed to vaporize 75.0 g of diethyl ether

Explanation:

First we have to calculate the moles of diethyl ether

$\text{Moles of diethyl ether}=\frac{\text{Mass of diethyl ether}}{\text{Molar mass of diethyl ether}}=\frac{75.0g}{74g/mole}=1.01moles$

As, 1 mole of diethyl ether require heat = 26.5 kJ

So, 1.01  moles of diethyl ether require heat = $\frac{26.5}{1}\times 1.01=26.8kJ$

Thus 26.8 kJ of energy is needed to vaporize 75.0 g of diethyl ether

Answer 2

The energy i.e. needed is 26.8 kJ of energy

• The calculation is as follows:

the moles of diethyl ether is

$= 75.0 g \div 74$

= 1.01 moles

Since one  mole of diethyl ether require heat = 26.8 kJ

Now the energy should be the same i.e. 26.8

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