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 `456 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.

Options

A. 73 g/mol
B. 50.8 g/mol
C. 37 g/mol
D. 41 g/mol

Explanation

To find the molecular weight of the organic liquid \( A \), we can follow these steps:

### Step 1: Understand the Given Information
- The boiling point of the mixture of liquid \( A \) and water is \( 90^\circ C \).
- The partial vapor pressure of water at this temperature is \( 456 \, \text{mm Hg} \).
- The atmospheric pressure is \( 736 \, \text{mm Hg} \).
- The weight ratio of liquid \( A \) to water is \( 2.5:1 \).

### Step 2: Calculate the Partial Vapor Pressure of Liquid \( A \)
Using Dalton's Law of Partial Pressures, we can find the partial vapor pressure of liquid \( A \):

\[ P_A = P_{total} - P_{water} \]

Where:
- \( P_{total} = 736 \, \text{mm Hg} \)
- \( P_{water} = 456 \, \text{mm Hg} \)

Substituting the values:

\[ P_A = 736 \, \text{mm Hg} - 456 \, \text{mm Hg} = 280 \, \text{mm Hg} \]

### Step 3: Set Up the Weight Ratio Equation
The weight ratio of the liquid \( A \) to water is given as \( 2.5:1 \). Let:
- \( w_1 = \text{weight of liquid } A \)
- \( w_2 = \text{weight of water} \)

From the weight ratio, we have:

\[ \frac{w_1}{w_2} = \frac{2.5}{1} \]

This implies:

\[ w_1 = 2.5 \, w_2 \]

### Step 4: Use the Formula Relating Molecular Weights and Partial Pressures
According to the formula:

\[ \frac{w_1}{w_2} = \frac{M_1 \cdot P_A}{M_2 \cdot P_{water}} \]

Where:
- \( M_1 = \text{molecular weight of liquid } A \)
- \( M_2 = \text{molecular weight of water} = 18 \, \text{g/mol} \)

Substituting the known values:

\[ \frac{2.5}{1} = \frac{M_1 \cdot 280}{18 \cdot 456} \]

### Step 5: Rearranging the Equation to Solve for \( M_1 \)
Rearranging the equation gives:

\[ M_1 = \frac{2.5 \cdot 18 \cdot 456}{280} \]

### Step 6: Calculate \( M_1 \)
Now, we can calculate \( M_1 \):

\[ M_1 = \frac{2.5 \cdot 18 \cdot 456}{280} = \frac{2.5 \cdot 18 \cdot 456}{280} \]

Calculating the numerator:

\[ 2.5 \cdot 18 = 45 \]
\[ 45 \cdot 456 = 20520 \]

Now divide by \( 280 \):

\[ M_1 = \frac{20520}{280} \approx 73.29 \, \text{g/mol} \]

### Conclusion
The molecular weight of the organic liquid \( A \) is approximately \( 73.29 \, \text{g/mol} \).