The vapour pressure of 2.1% solution of a non- electrolyte in water at `100^(@) C` is 755 mm Hg. Calculate the molar mass of the solute .

Explanation

To find the molar mass of the solute in the given problem, we can follow these steps:

### Step 1: Identify the given data
- Vapor pressure of pure water at 100°C, \( P_0 = 760 \, \text{mmHg} \)
- Vapor pressure of the solution, \( P = 755 \, \text{mmHg} \)
- Mass percent of the solute, \( \text{mass percent} = 2.1\% \)

### Step 2: Calculate the change in vapor pressure
Using the formula for the change in vapor pressure:
\[ \Delta P = P_0 - P = 760 \, \text{mmHg} - 755 \, \text{mmHg} = 5 \, \text{mmHg} \]

### Step 3: Write the formula from Raoult's Law
According to Raoult's Law:
\[ \frac{\Delta P}{P_0} = \frac{W_2}{W_1} \cdot \frac{M_1}{M_2} \]
Where:
- \( W_2 \) = mass of the solute
- \( W_1 \) = mass of the solvent
- \( M_1 \) = molar mass of the solvent (water, \( M_1 = 18 \, \text{g/mol} \))
- \( M_2 \) = molar mass of the solute (which we need to find)

### Step 4: Substitute the known values
From the mass percent, we know that:
- Mass of solute, \( W_2 = 2.1 \, \text{g} \)
- Mass of solvent, \( W_1 = 100 \, \text{g} - 2.1 \, \text{g} = 97.9 \, \text{g} \)

Now substituting the values into the equation:
\[ \frac{5}{760} = \frac{2.1}{97.9} \cdot \frac{18}{M_2} \]

### Step 5: Cross-multiply to solve for \( M_2 \)
Cross-multiplying gives:
\[ 5 \cdot M_2 = \frac{2.1 \cdot 18 \cdot 760}{97.9} \]

### Step 6: Calculate the right side
Calculating the right side:
\[ 5 \cdot M_2 = \frac{2.1 \cdot 18 \cdot 760}{97.9} \approx \frac{28861.2}{97.9} \approx 294.2 \]

### Step 7: Solve for \( M_2 \)
Now, divide both sides by 5:
\[ M_2 = \frac{294.2}{5} \approx 58.84 \, \text{g/mol} \]

### Conclusion
The molar mass of the solute is approximately \( 58.84 \, \text{g/mol} \).

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