An organic liquid, `A` is immiscible with water. When boiled together with water, the boiling point is `90^(@)C` which the partial vapour pressure of water is `526 mm Hg`. The superincumbent (atmospheric) pressure is `736 mm Hg`. The weight ratio of the liqid water collected `2.5:1`. What is the molecular weight of the liquid.

Answer

To find the molecular weight of the organic liquid `A`, we can use the relationship between the weight of the liquid and water collected, the partial vapor pressures, and their molecular weights. Here’s a step-by-step solution:

### Step 1: Understand the Given Data
- The boiling point of the mixture is 90°C.
- The partial vapor pressure of water (P_w) = 526 mm Hg.
- The atmospheric pressure (P_total) = 736 mm Hg.
- The weight ratio of the liquid to water (w1:w2) = 2.5:1.

### Step 2: Calculate the Partial Vapor Pressure of Liquid `A`
Using the formula for total pressure:
\[ P_{total} = P_w + P_A \]
Where \( P_A \) is the partial vapor pressure of liquid `A`.

Rearranging gives:
\[ P_A = P_{total} - P_w = 736 \text{ mm Hg} - 526 \text{ mm Hg} = 210 \text{ mm Hg} \]

### Step 3: Set Up the Weight Ratio Equation
From the weight ratio given:
\[ \frac{w_1}{w_2} = 2.5 \]
Let \( w_1 = 2.5x \) and \( w_2 = x \).

### Step 4: Use the Formula Relating Weights and Vapor Pressures
The relationship between the weights and vapor pressures is given by:
\[ \frac{w_1}{w_2} = \frac{M_A \cdot P_A}{M_w \cdot P_w} \]
Where:
- \( M_A \) = molecular weight of liquid `A`
- \( M_w \) = molecular weight of water = 18 g/mol
- \( P_A \) = partial vapor pressure of liquid `A` = 210 mm Hg
- \( P_w \) = partial vapor pressure of water = 526 mm Hg

Substituting the values:
\[ 2.5 = \frac{M_A \cdot 210}{18 \cdot 526} \]

### Step 5: Rearranging to Solve for Molecular Weight \( M_A \)
Rearranging the equation gives:
\[ M_A = 2.5 \cdot \frac{18 \cdot 526}{210} \]

### Step 6: Calculate \( M_A \)
Calculating the right side:
1. Calculate \( 18 \cdot 526 = 9488 \).
2. Now divide by 210: \( \frac{9488}{210} \approx 45.06 \).
3. Finally, multiply by 2.5: \( 2.5 \cdot 45.06 \approx 112.65 \).

Thus, the molecular weight of the liquid `A` is approximately:
\[ M_A \approx 112.7 \text{ g/mol} \]

### Final Answer
The molecular weight of the organic liquid `A` is approximately **112.7 g/mol**.

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