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  • The vapour pressure of a solution containing 2 moles of ethanol  <img alt="\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cleft%28%5

The vapour pressure of a solution containing 2 moles of ethanol  \left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right) and 1 mole of acetone \left(\mathrm{CH}_3 \mathrm{COCH}_3\right) is found to be 93.3 \mathrm{kPa} at 298 \mathrm{~K}. If the vapour pressure of pure ethanol and pure acetone at this temperature are 100 \, \, \mathrm{kPa} and 75\, \, \mathrm{kPa} respectively, what is the vapour pressure of an ideal solution of the same composition?

Option: 1

90.5 \, \, \mathrm{kPa}


Option: 2

103.5 \, \, \mathrm{kPa}


Option: 3

94.7 \, \, \mathrm{kPa}


Option: 4

96.1 \, \, \mathrm{kPa}


Answers (1)

best_answer

The vapour pressure of a solution containing multiple components can be calculated using Raoult's law as follows:

\begin{aligned} & P_A=X_A P_A^{\circ} \\ & P_B=X_B P_B^{\circ} \end{aligned}


where P_A and P_B are the partial pressures of the components A and B respectively,   X_A and X_B are their mole fractions in the solution, and P_A^{\circ} and P_B^{\circ} are their vapour pressures in the pure state.

In the case of a non-ideal solution showing negative deviation from Raoult's law, the vapour pressure of the solution is lower than that predicted by Raoult's law due to the stronger intermolecular forces between unlike molecules. Therefore, we can calculate the vapour pressure of an ideal solution of the same composition using Raoult's law by assuming the solution shows ideal behaviour:

P_{\text {ideal }}=X_A P_A^{\circ}+X_B P_B^{\circ}

Here, X_A=2 / 3 \text { and } X_B=1 / 3, P_A^{\circ}=100 \mathrm{kPa} \text {, and } P_B^{\circ}=75 \mathrm{kPa} \text {. }

P_{\text {ideal }}=(2 / 3)(100 k P a)+(1 / 3)(75 k P a)=83.33 k P a

However, the given solution is non-ideal and shows negative deviation from Raoult's law, resulting in a lower vapour pressure of 93.3 kPa.

To find the vapour pressure of an ideal solution with the same composition as the given solution, we can use the equation for the degree of non-ideality, i.e.,

\begin{aligned} & \Delta P=P_{\text {observed }}-P_{\text {ideal }} \\ & \Delta P=93.3 \mathrm{kPa}-83.33 \mathrm{kPa}=9.97 \mathrm{kPa} \end{aligned}

Since the observed vapour pressure is lower than that of an ideal solution, the deviation is negative. Therefore, the vapour pressure of an ideal solution with the same composition as the given solution can be calculated as:

\begin{aligned} & P_{\text {ideal }}=P_{\text {observed }}+\Delta P \\ & P_{\text {ideal }}=93.3 \mathrm{kPa}+9.97 \mathrm{kPa}=103.27 \mathrm{kPa} \end{aligned}

Therefore, the answer is option B,103.27 \, \, \mathrm{kPa} , which is closest to the calculated value of 103.5 \, \, \mathrm{kPa}.

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