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 `345 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. 25.7 g/mol
B. 39.7 g/mol
C. 26 g/mol
D. 20 g/mol

Answer

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

### Step 1: Understand the given data
- The boiling point of the mixture is \(90^\circ C\).
- The partial vapor pressure of water at this temperature is \(P_{H_2O} = 345 \, \text{mm Hg}\).
- The atmospheric pressure is \(P_{total} = 736 \, \text{mm Hg}\).
- The weight ratio of the liquid to water is \(W_A : W_{H_2O} = 2.5 : 1\).

### Step 2: Calculate the partial vapor pressure of the organic liquid
Using Dalton's law of partial pressures:
\[ P_A = P_{total} - P_{H_2O} = 736 \, \text{mm Hg} - 345 \, \text{mm Hg} = 391 \, \text{mm Hg} \]

### Step 3: Set up the equation using the weight ratio
From the weight ratio, we have:
\[ \frac{W_A}{W_{H_2O}} = \frac{M_A \cdot P_A}{M_{H_2O} \cdot P_{H_2O}} \]
Where:
- \(W_A\) = weight of the organic liquid
- \(W_{H_2O}\) = weight of water
- \(M_A\) = molecular weight of the organic liquid (unknown)
- \(M_{H_2O} = 18 \, \text{g/mol}\) (known)
- \(P_A = 391 \, \text{mm Hg}\) (calculated)
- \(P_{H_2O} = 345 \, \text{mm Hg}\) (given)

### Step 4: Substitute the known values into the equation
Given the weight ratio \(W_A : W_{H_2O} = 2.5 : 1\), we can express this as:
\[ \frac{W_A}{W_{H_2O}} = 2.5 \]
Thus, we can write:
\[ 2.5 = \frac{M_A \cdot 391}{18 \cdot 345} \]

### Step 5: Rearrange the equation to solve for \(M_A\)
Rearranging gives:
\[ M_A = \frac{2.5 \cdot 18 \cdot 345}{391} \]

### Step 6: Calculate \(M_A\)
Now, we can calculate \(M_A\):
\[ M_A = \frac{2.5 \cdot 18 \cdot 345}{391} \approx \frac{15450}{391} \approx 39.5 \, \text{g/mol} \]

### Conclusion
The molecular weight of the organic liquid `A` is approximately \(39.5 \, \text{g/mol}\).

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