1. Use Raoult’s law for an ideal binary solution

For a mixture of ethyl alcohol and propyl alcohol, $P_{\text{total}} = x_1 P_1^0 + x_2 P_2^0$ where:

• $x_1 = 0.6$ = mole fraction of ethyl alcohol
• $x_2 = 1 - 0.6 = 0.4$ = mole fraction of propyl alcohol
• $P_{\text{total}} = 290\ \text{mm}$
• $P_2^0 = 200\ \text{mm}$
• $P_1^0 =$ vapour pressure of pure ethyl alcohol (to be found)

2. Substitute the given values

$290 = (0.6)P_1^0 + (0.4)(200)$

3. Simplify

$290 = 0.6P_1^0 + 80$

$290 - 80 = 0.6P_1^0$

$210 = 0.6P_1^0$

4. Solve for $P_1^0$

$P_1^0 = \frac{210}{0.6} = 350\ \text{mm}$

5. Match with the options

$350\ \text{mm}$ corresponds to Option B.

Final Answer

The vapour pressure of ethyl alcohol at $300\ \text{K}$ is: $\boxed{350\ \text{mm}}$ So, the correct option is B.

6. Comparison with stored correct answer

Stored correct answer = B

My derived answer = B

Hence, they agree.