Question

iclicker: how much energy is required to vaporize 32.9 g of ethanol at its boiling point, if its $delta h_{vap}$ is 40.7 kj mol$^{-1}$? 1a) 29.0 kj 2b) 3.83 kj 3c) 6.30 kj 4d) 17.6 kj 5e) 7.60 kj your answer no response incorrect correct answer a all results a 62% b 12% c 10% d 6% e 6%

Explanation:

Step1: Calculate moles of ethanol

The molar mass of ethanol ($C_2H_5OH$) is $M=(2\times12.01 + 6\times1.01+16.00)\text{ g/mol}=46.08\text{ g/mol}$. The number of moles $n$ of ethanol with a mass $m = 32.9\text{ g}$ is $n=\frac{m}{M}=\frac{32.9\text{ g}}{46.08\text{ g/mol}}\approx0.714\text{ mol}$.

Step2: Calculate energy required

The heat of vaporization $\Delta H_{vap}=40.7\text{ kJ/mol}$. The energy $q$ required for vaporization is $q = n\times\Delta H_{vap}$. Substituting $n = 0.714\text{ mol}$ and $\Delta H_{vap}=40.7\text{ kJ/mol}$, we get $q=0.714\text{ mol}\times40.7\text{ kJ/mol}\approx29.0\text{ kJ}$.

Answer:

A. 29.0 kJ