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Step-by-Step Solution
Step 1: Identify the components and data
• Let ethyl alcohol be component A, and propyl alcohol be component B.
• Given total vapor pressure of the mixture at 300 K: $P = 290 \text{ mm Hg}$.
• Mole fraction of ethyl alcohol, $x_A = 0.6$.
• Accordingly, mole fraction of propyl alcohol, $x_B = 1 - x_A = 0.4$.
• Vapor pressure of pure propyl alcohol (component B): $P_B^\circ = 200 \text{ mm Hg}$.
• We need to find the vapor pressure of pure ethyl alcohol (component A), $P_A^\circ$.
Step 2: Write the expression for total vapor pressure
For an ideal solution, the total vapor pressure is given by Raoult’s Law:
$P_{\text{total}} = x_A \, P_A^\circ + x_B \, P_B^\circ$
Step 3: Substitute the known values
Substitute $P_{\text{total}} = 290$, $x_A = 0.6$, $x_B = 0.4$, and $P_B^\circ = 200$:
$290 = (0.6)\,P_A^\circ + (0.4)\,(200)$
Step 4: Solve for $P_A^\circ$
First calculate the contribution from propyl alcohol:
$(0.4) \times 200 = 80$
So the equation becomes
$290 = 0.6\,P_A^\circ + 80$
Rearrange to isolate $P_A^\circ$:
$290 - 80 = 0.6\,P_A^\circ$
$210 = 0.6\,P_A^\circ$
$P_A^\circ = \frac{210}{0.6} = 350 \text{ mm Hg}$
Step 5: Conclude the result
Hence, the vapor pressure of pure ethyl alcohol at 300 K is $350 \text{ mm Hg}$.
