At `40^(@)C`, the vapour pressure in torr of methyl and ethyl alcohol solutions is represented by `P = 119 X_(A)+135`, where `X_(A)` is mole fraction of methyl alcohol. The value of `(P_(B)^(@))/(X_(B))` at lim `X_(A) rarr 0)`, and `(P_(A)^(@))/(X_(A))` at lim `X_(B) rarr 0` are:

Options

A. (a) `135, 254`
B. (b) `135, 230`
C. ( c) `119, 135`
D. (d) `140, 135`

Explanation

To solve the problem, we need to analyze the given equation for the vapor pressure of a solution of methyl and ethyl alcohol. The equation provided is:

\[ P = 119 X_A + 135 \]

where \( P \) is the vapor pressure, and \( X_A \) is the mole fraction of methyl alcohol.

### Step 1: Find \( P_B^0 / X_B \) as \( X_A \to 0 \)

When \( X_A \) approaches 0, the mole fraction of ethyl alcohol \( X_B \) approaches 1 (since \( X_A + X_B = 1 \)). Therefore, we can substitute \( X_A = 0 \) into the equation:

\[ P = 119(0) + 135 = 135 \text{ torr} \]

Now, since \( P_B^0 \) is the vapor pressure of pure ethyl alcohol, we can express it as:

\[ P_B^0 = P \text{ when } X_A \to 0 = 135 \text{ torr} \]

Now, we can find \( P_B^0 / X_B \):

\[ \frac{P_B^0}{X_B} = \frac{135}{1} = 135 \]

### Step 2: Find \( P_A^0 / X_A \) as \( X_B \to 0 \)

When \( X_B \) approaches 0, the mole fraction of methyl alcohol \( X_A \) approaches 1. Therefore, we can substitute \( X_B = 0 \) into the equation:

\[ P = 119(1) + 135 = 119 + 135 = 254 \text{ torr} \]

Now, since \( P_A^0 \) is the vapor pressure of pure methyl alcohol, we can express it as:

\[ P_A^0 = P \text{ when } X_B \to 0 = 254 \text{ torr} \]

Now, we can find \( P_A^0 / X_A \):

\[ \frac{P_A^0}{X_A} = \frac{254}{1} = 254 \]

### Final Answers

Thus, the values are:
- \( \frac{P_B^0}{X_B} \) as \( X_A \to 0 \) = 135
- \( \frac{P_A^0}{X_A} \) as \( X_B \to 0 \) = 254

### Summary of Results
- \( \frac{P_B^0}{X_B} = 135 \)
- \( \frac{P_A^0}{X_A} = 254 \)