When a liquid that is immiscible with water was steam distilled at `95.2^(@)C` at a total pressure of 748 torr, the distilled contained 1.25 g of the liquid per gram of water . The vapour pressure of water is 648 torr at `95.2^(@)C`, what is the molar mass of liquid?

Options

A. 7.975 g/mol
B. 166 g/ mol
C. 145.8 g/mol
D. None of these

Answer

To solve the problem of finding the molar mass of the immiscible liquid distilled with water, we can follow these steps:

### Step 1: Gather the Given Data
- Weight of the liquid (WL) = 1.25 g
- Weight of water (WH2O) = 1 g
- Total pressure (PT) = 748 torr
- Vapor pressure of water (PH2O) = 648 torr

### Step 2: Calculate the Vapor Pressure of the Liquid (PL)
Using the formula:
\[ PL = PT - PH2O \]
Substituting the values:
\[ PL = 748 \, \text{torr} - 648 \, \text{torr} = 100 \, \text{torr} \]

### Step 3: Apply Raoult's Law
According to Raoult's Law:
\[ \frac{PL}{PH2O} = \frac{n_{solute}}{n_{solvent}} \]
Where:
- \( n_{solute} \) = number of moles of the liquid
- \( n_{solvent} \) = number of moles of water

### Step 4: Calculate the Number of Moles of Water
The molar mass of water (H2O) is approximately 18 g/mol.
Thus, the number of moles of water (nH2O) is:
\[ n_{H2O} = \frac{WH2O}{M_{H2O}} = \frac{1 \, \text{g}}{18 \, \text{g/mol}} = \frac{1}{18} \, \text{mol} \]

### Step 5: Set Up the Equation
From Raoult's Law:
\[ \frac{100 \, \text{torr}}{648 \, \text{torr}} = \frac{n_{solute}}{n_{H2O}} \]
Substituting for nH2O:
\[ \frac{100}{648} = \frac{n_{solute}}{\frac{1}{18}} \]

### Step 6: Solve for nsolute
Rearranging gives:
\[ n_{solute} = \frac{100}{648} \times \frac{1}{18} \]
Calculating this:
\[ n_{solute} = \frac{100}{648 \times 18} \]

### Step 7: Calculate the Molar Mass of the Liquid
The number of moles of the liquid (nsolute) can also be expressed as:
\[ n_{solute} = \frac{WL}{M_{L}} \]
Where \( WL = 1.25 \, \text{g} \) and \( M_{L} \) is the molar mass of the liquid.

Setting the two expressions for nsolute equal:
\[ \frac{WL}{M_{L}} = \frac{100}{648 \times 18} \]
Substituting WL:
\[ \frac{1.25}{M_{L}} = \frac{100}{648 \times 18} \]

### Step 8: Solve for ML
Rearranging gives:
\[ M_{L} = \frac{1.25 \times 648 \times 18}{100} \]
Calculating this:
\[ M_{L} = \frac{1.25 \times 648 \times 18}{100} = 145.8 \, \text{g/mol} \]

### Final Answer
The molar mass of the liquid is approximately **145.8 g/mol**.

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