10g of a certain non-volatile solute was dissolved in 100g of water at `20^(@)C`. The vapour pressure was lowered in 100g of water at `20^(@)C`. The vapour pressure was lowered from 17.3555 to 17.235 mm. Calculate the molecular mass of the solute.

Answer

To calculate the molecular mass of the solute, we will follow these steps:

### Step 1: Determine the change in vapor pressure
The initial vapor pressure of pure water (P₀) is given as 17.355 mm, and the vapor pressure of the solution (P) is 17.235 mm.

Change in vapor pressure (ΔP) can be calculated as:
\[ \Delta P = P_0 - P = 17.355 \, \text{mm} - 17.235 \, \text{mm} = 0.120 \, \text{mm} \]

### Step 2: Calculate the relative lowering of vapor pressure
The relative lowering of vapor pressure (ΔP/P₀) is given by:
\[ \text{Relative lowering} = \frac{\Delta P}{P_0} = \frac{0.120 \, \text{mm}}{17.355 \, \text{mm}} \approx 0.0069 \]

### Step 3: Relate the relative lowering to the mole fraction of the solute
According to Raoult's Law, the relative lowering of vapor pressure is equal to the mole fraction of the solute (X_solute):
\[ \text{Relative lowering} = X_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} \]

### Step 4: Calculate the number of moles of solvent (water)
The number of moles of water (n_solvent) can be calculated using the formula:
\[ n_{\text{solvent}} = \frac{\text{mass of solvent}}{\text{molar mass of solvent}} = \frac{100 \, \text{g}}{18 \, \text{g/mol}} \approx 5.56 \, \text{mol} \]

### Step 5: Set up the equation for mole fraction
Let the number of moles of the solute be \( n_{\text{solute}} = \frac{10 \, \text{g}}{M} \) where M is the molar mass of the solute. The mole fraction can be expressed as:
\[ X_{\text{solute}} = \frac{\frac{10}{M}}{\frac{10}{M} + 5.56} \]

### Step 6: Substitute the mole fraction into the relative lowering equation
We can substitute the mole fraction into the relative lowering equation:
\[ 0.0069 = \frac{\frac{10}{M}}{\frac{10}{M} + 5.56} \]

### Step 7: Cross-multiply and solve for M
Cross-multiplying gives:
\[ 0.0069 \left(\frac{10}{M} + 5.56\right) = \frac{10}{M} \]
Expanding and rearranging:
\[ 0.0069 \cdot 5.56 = \frac{10}{M} - 0.0069 \cdot \frac{10}{M} \]
\[ 0.0384 = \frac{10(1 - 0.0069)}{M} \]
\[ 0.0384 = \frac{10 \cdot 0.9931}{M} \]
\[ M = \frac{10 \cdot 0.9931}{0.0384} \approx 258.5 \, \text{g/mol} \]

### Final Answer
The molecular mass of the solute is approximately **258.5 g/mol**.

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