Respuesta :
Answer:
The amount of energy required will be 86.7 kJ.
Explanation:
Mass of ethanol = 98.6 g
Moles of ethanol = [tex]\frac{98 g}{46 g/mol}=2.1434 moles[/tex]
Enthalpy of vaporization of ethanol =[tex]\Delta H_{vap}=40.5 kJ/mol[/tex]
To vaporize 1 mol of ethanol we need 40.5 kJ of energy.
Then for 2.1434 moles the energy required will be:
[tex]40.5 kJ\times 2.1432 kJ=86.79 kJ[/tex]
The closest answer from the given options is 86.7 kJ.
86.7 kJ
Further explanation
Given:
Molar mass (Mr) of ethanol = 46.07 g/mol
[tex]\Delta H_{vap} \ C_2H_5OH = 40.5 \ kJ/mol[/tex]
Question:
How much energy is required to vaporize 98.6 g of ethanol (C₂H₅OH) at its boiling point?
The Process:
Observe this [tex]\Delta H_{vap} \ C_2H_5OH = 40.5 \ kJ/mol[/tex]. This means that 40.5 kJ of energy is required to vaporize every 1 mole of ethanol at its boiling point. Therefore we must first convert grams into moles.
Question:
Step-1:
Let us count the number of moles of 98.6 g of ethanol.
[tex]\boxed{moles \ (n) = \frac{mass}{Mr}}[/tex]
[tex]\boxed{moles \ (n) = \frac{98.6 \ g}{46.07 \ g/mol}}[/tex]
We obtain 2.14 moles of ethanol.
Step-2:
Let us calculate how much energy is required to vaporize 2.14 moles of ethanol at its boiling point.
The amount of energy required is [tex]\boxed{n \times \Delta H_{vap}}[/tex]
[tex]\boxed{ \ The \ energy = 2.14 \ moles \times 40.5 \ \frac{kJ}{mol} \ }[/tex]
[tex]\boxed{ \ The \ energy = 86.67 \ kJ \approx 86.7 \ kJ \ }[/tex]
Thus, the amount of energy required to vaporize 98.6 g of ethanol at its boiling point is 86.7 kJ.
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Quick Steps
[tex]\boxed{ \ The \ energy \ required = 98.6 \ g \times \frac{1 \ mol}{46.07 \ g} \times 40.5 \ \frac{kJ}{mol} \approx 86.7 \ kJ \ }[/tex]
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Keywords: how much, energy required, to vaporize, ethanol, C₂H₅OH, its boiling point, ΔHvap, molar mass, moles, converts, kJ/mol
