At `20%(@)C` the vapour pressure of ether is 442 mmHg. When `6.1`g of a substance is dissolved in 50g of ether (mo.wt.74) the vapour pressure falls to 410mm. What is the molecular weight of the substance ?

Answer

To find the molecular weight of the substance dissolved in ether, we can follow these steps:

### Step 1: Calculate the decrease in vapor pressure
The initial vapor pressure of ether is 442 mmHg, and the final vapor pressure after dissolving the substance is 410 mmHg.

\[ \text{Decrease in vapor pressure} = P_0 - P_f = 442 \, \text{mmHg} - 410 \, \text{mmHg} = 32 \, \text{mmHg} \]

### Step 2: Use Raoult's Law
According to Raoult's Law, the decrease in vapor pressure is related to the mole fraction of the solute:

\[ \Delta P = P_0 \cdot X_{\text{solute}} \]

Where:
- \( \Delta P \) is the decrease in vapor pressure (32 mmHg)
- \( P_0 \) is the initial vapor pressure (442 mmHg)
- \( X_{\text{solute}} \) is the mole fraction of the solute

Rearranging gives:

\[ X_{\text{solute}} = \frac{\Delta P}{P_0} = \frac{32}{442} \]

Calculating this gives:

\[ X_{\text{solute}} \approx 0.0724 \]

### Step 3: Calculate the mole fraction of the solvent (ether)
The mole fraction of the solvent (ether) can be calculated as:

\[ X_{\text{solvent}} = 1 - X_{\text{solute}} = 1 - 0.0724 \approx 0.9276 \]

### Step 4: Relate mole fraction to moles
The mole fraction is also defined as:

\[ X_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} \]

Where:
- \( n_{\text{solute}} \) is the number of moles of the solute
- \( n_{\text{solvent}} \) is the number of moles of the solvent (ether)

### Step 5: Calculate moles of ether
The number of moles of ether can be calculated using its mass and molar mass:

\[ n_{\text{solvent}} = \frac{\text{mass}}{\text{molar mass}} = \frac{50 \, \text{g}}{74 \, \text{g/mol}} \approx 0.6757 \, \text{mol} \]

### Step 6: Set up the equation
Now we can set up the equation using the mole fraction:

\[ 0.0724 = \frac{n_{\text{solute}}}{n_{\text{solute}} + 0.6757} \]

### Step 7: Substitute \( n_{\text{solute}} \)
The number of moles of the solute can be expressed as:

\[ n_{\text{solute}} = \frac{6.1 \, \text{g}}{M} \]

Where \( M \) is the molar mass of the solute (unknown). Substituting this into the equation gives:

\[ 0.0724 = \frac{\frac{6.1}{M}}{\frac{6.1}{M} + 0.6757} \]

### Step 8: Cross-multiply and solve for \( M \)
Cross-multiplying gives:

\[ 0.0724 \left( \frac{6.1}{M} + 0.6757 \right) = \frac{6.1}{M} \]

Expanding and simplifying leads to:

\[ 0.0724 \cdot 0.6757 = \frac{6.1}{M} - 0.0724 \cdot \frac{6.1}{M} \]

Factoring out \( \frac{6.1}{M} \):

\[ 0.0724 \cdot 0.6757 = \frac{6.1}{M} (1 - 0.0724) \]

Calculating \( 0.0724 \cdot 0.6757 \):

\[ 0.0724 \cdot 0.6757 \approx 0.0489 \]

And \( (1 - 0.0724) \approx 0.9276 \):

\[ 0.0489 = \frac{6.1}{M} \cdot 0.9276 \]

### Step 9: Solve for \( M \)
Rearranging gives:

\[ M = \frac{6.1 \cdot 0.9276}{0.0489} \]

Calculating this gives:

\[ M \approx 124.6 \, \text{g/mol} \]

### Final Answer
The molecular weight of the substance is approximately **124.6 g/mol**.

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