In an experiment , 1804 g of mannitol were dissolved in 100 g of water . The vapour pressure of water was lowered by 0.309 mm Hg from 17.535 mm Hg. Calculate the molar mass of mannitol .

Explanation

To calculate the molar mass of mannitol based on the provided data, we can follow these steps:

### Step 1: Identify the Given Data
- Mass of mannitol (solute), \( w_2 = 1804 \, \text{g} \)
- Mass of water (solvent), \( w_1 = 100 \, \text{g} \)
- Lowering in vapor pressure, \( P_0 - P = 0.309 \, \text{mm Hg} \)
- Vapor pressure of pure water, \( P_0 = 17.535 \, \text{mm Hg} \)
- Molar mass of water, \( M_1 = 18 \, \text{g/mol} \)

### Step 2: Use the Formula for Vapor Pressure Lowering
The formula relating the lowering of vapor pressure to the molar mass of the solute is given by:

\[ \frac{P_0 - P}{P_0} = \frac{w_2}{w_1} \cdot \frac{M_1}{M_2} \]

Where:
- \( P_0 \) = vapor pressure of pure solvent
- \( P \) = vapor pressure of the solution
- \( w_2 \) = mass of solute
- \( w_1 \) = mass of solvent
- \( M_1 \) = molar mass of solvent
- \( M_2 \) = molar mass of solute (mannitol)

### Step 3: Rearranging the Formula
We need to rearrange the formula to solve for the molar mass of mannitol (\( M_2 \)):

\[ M_2 = \frac{w_2}{w_1} \cdot \frac{M_1 \cdot P_0}{P_0 - P} \]

### Step 4: Substitute the Values
Now, substitute the known values into the rearranged formula:

\[ M_2 = \frac{1804 \, \text{g}}{100 \, \text{g}} \cdot \frac{18 \, \text{g/mol} \cdot 17.535 \, \text{mm Hg}}{0.309 \, \text{mm Hg}} \]

### Step 5: Calculate Each Part
1. Calculate \( \frac{1804}{100} = 18.04 \)
2. Calculate \( 18 \cdot 17.535 = 315.63 \)
3. Finally, calculate \( \frac{315.63}{0.309} \approx 1019.77 \)

### Step 6: Final Calculation
Now, multiply the results from steps 1 and 3:

\[ M_2 = 18.04 \cdot 1019.77 \approx 183.68 \, \text{g/mol} \]

### Conclusion
The molar mass of mannitol is approximately \( 183.68 \, \text{g/mol} \).

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