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 `652 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 solve the problem step by step, we will use the given information and apply the relevant formulas.

### Step 1: Understand the Given Data
- The boiling point of the mixture is \(90^\circ C\).
- The partial vapor pressure of water (\(P_w\)) is \(652 \, \text{mm Hg}\).
- The atmospheric pressure (\(P_{atm}\)) is \(736 \, \text{mm Hg}\).
- The weight ratio of the organic liquid \(A\) to water is \(2.5:1\).

### Step 2: Calculate the Partial Vapor Pressure of Liquid \(A\)
Using the formula for total pressure:
\[ P_{atm} = P_w + P_A \]
Where \(P_A\) is the partial vapor pressure of liquid \(A\). Rearranging gives:
\[ P_A = P_{atm} - P_w = 736 \, \text{mm Hg} - 652 \, \text{mm Hg} = 84 \, \text{mm Hg} \]

### Step 3: Set Up the Weight Ratio Equation
The weight ratio of the two liquids can be expressed as:
\[ \frac{w_1}{w_2} = \frac{m_1 \cdot P_A}{m_2 \cdot P_w} \]
Where:
- \(w_1\) = weight of liquid \(A\)
- \(w_2\) = weight of water
- \(m_1\) = molecular weight of liquid \(A\) (unknown)
- \(m_2\) = molecular weight of water = \(18 \, \text{g/mol}\)
- \(P_A\) = partial vapor pressure of liquid \(A\)
- \(P_w\) = partial vapor pressure of water

Given \(w_1:w_2 = 2.5:1\), we can write:
\[ \frac{w_1}{w_2} = 2.5 \]
Thus, substituting into the equation:
\[ 2.5 = \frac{m_1 \cdot 84}{18 \cdot 652} \]

### Step 4: Rearranging the Equation
Rearranging the equation to solve for \(m_1\):
\[ m_1 = 2.5 \cdot \frac{18 \cdot 652}{84} \]

### Step 5: Calculate the Molecular Weight of Liquid \(A\)
Now, we can calculate \(m_1\):
\[ m_1 = 2.5 \cdot \frac{18 \cdot 652}{84} \]
Calculating the values:
\[ m_1 = 2.5 \cdot \frac{11736}{84} = 2.5 \cdot 139.7143 \approx 349.2857 \, \text{g/mol} \]

### Final Answer
The molecular weight of the organic liquid \(A\) is approximately \(349.3 \, \text{g/mol}\).

---