Question

a solution is made by mixing 38.0 ml of ethanol, $ce{c_{2}h_{6}o}$, and 62.0 ml of water. assuming ideal behavior, what is the vapor pressure of the solution at 20 °c? the relevant values at 20 °c are included in the table.

liquid

density (g/ml)

vapor pressure, $p^\circ$ (torr)

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ethanol

0.789

43.9

water

0.998

17.5

vapor pressure: _______ torr

Explanation:

Step1: Calculate mass of ethanol

$m_{\text{ethanol}} = \text{Volume} \times \text{Density} = 38.0\ \text{mL} \times 0.789\ \text{g/mL} = 29.982\ \text{g}$

Step2: Calculate mass of water

$m_{\text{water}} = \text{Volume} \times \text{Density} = 62.0\ \text{mL} \times 0.998\ \text{g/mL} = 61.876\ \text{g}$

Step3: Calculate moles of ethanol

Molar mass of $\text{C}_2\text{H}_6\text{O} = (2\times12.01)+(6\times1.008)+16.00 = 46.07\ \text{g/mol}$
$n_{\text{ethanol}} = \frac{m_{\text{ethanol}}}{\text{Molar Mass}} = \frac{29.982\ \text{g}}{46.07\ \text{g/mol}} \approx 0.6508\ \text{mol}$

Step4: Calculate moles of water

Molar mass of $\text{H}_2\text{O} = (2\times1.008)+16.00 = 18.016\ \text{g/mol}$
$n_{\text{water}} = \frac{m_{\text{water}}}{\text{Molar Mass}} = \frac{61.876\ \text{g}}{18.016\ \text{g/mol}} \approx 3.4345\ \text{mol}$

Step5: Calculate total moles

$n_{\text{total}} = n_{\text{ethanol}} + n_{\text{water}} = 0.6508\ \text{mol} + 3.4345\ \text{mol} \approx 4.0853\ \text{mol}$

Step6: Calculate mole fractions

$x_{\text{ethanol}} = \frac{n_{\text{ethanol}}}{n_{\text{total}}} = \frac{0.6508}{4.0853} \approx 0.1593$
$x_{\text{water}} = \frac{n_{\text{water}}}{n_{\text{total}}} = \frac{3.4345}{4.0853} \approx 0.8407$

Step7: Apply Raoult's Law

$P_{\text{solution}} = x_{\text{ethanol}}P^\circ_{\text{ethanol}} + x_{\text{water}}P^\circ_{\text{water}}$
$P_{\text{solution}} = (0.1593 \times 43.9\ \text{Torr}) + (0.8407 \times 17.5\ \text{Torr})$
$P_{\text{solution}} \approx 6.993\ \text{Torr} + 14.712\ \text{Torr} \approx 21.705\ \text{Torr}$

Answer:

21.7 Torr