How much energy is required to vaporize 98.6 g of ethanol (C2H5OH) at its boiling point, if its ΔHvap is 40.5 kJ/mol? How much energy is required to vaporize 98.6 g of ethanol (C2H5OH) at its boiling point, if its ΔHvap is 40.5 kJ/mol? 52.8 kJ 11.5 kJ 86.7 kJ 39.9 kJ 18.9 kJ

QUESTION POSTED AT 18/04/2020 - 06:20 PM

Answered by admin AT 18/04/2020 - 06:20 PM

86.7 kJ

Further explanation

Given:

Molar mass (Mr) of ethanol = 46.07 g/mol

\Delta H_{vap} \ C_2H_5OH = 40.5 \ kJ/mol

Question:

How much energy is required to vaporize 98.6 g of ethanol (C₂H₅OH) at its boiling point?

The Process:

Observe this \Delta H_{vap} \ C_2H_5OH = 40.5 \ kJ/mol. 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.

\boxed{moles \ (n) = \frac{mass}{Mr}}

\boxed{moles \ (n) = \frac{98.6 \ g}{46.07 \ g/mol}}

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 \boxed{n \times \Delta H_{vap}}

\boxed{ \ The \ energy = 2.14 \ moles \times 40.5 \ \frac{kJ}{mol} \ }

\boxed{ \ The \ energy = 86.67 \ kJ \approx 86.7 \ kJ \ }

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

\boxed{ \ The \ energy \ required = 98.6 \ g \times \frac{1 \ mol}{46.07 \ g} \times 40.5 \ \frac{kJ}{mol} \approx 86.7 \ kJ \ }

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Keywords: how much, energy required, to vaporize, ethanol, C₂H₅OH, its boiling point, ΔHvap, molar mass, moles, converts, kJ/mol

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