A solution of an organic compound is prepared by dissolving 34.2 g in 500 g of water.
Calculate the molar mass of the compound and freezing point of the solution.
Given that `K_(b)` for water `="0.52 K mol"^(-1)` B.pt of solution `=100.104^(@)C.K_(f)` for water `=1.87"K mol"^(-1)`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the elevation in boiling point (ΔT_B)
The boiling point of the solution (T_B) is given as 100.104 °C, and the boiling point of pure water (T_B₀) is 100 °C.

\[ \Delta T_B = T_B - T_B₀ = 100.104 °C - 100 °C = 0.104 °C \]

### Step 2: Use the formula for boiling point elevation
The formula for boiling point elevation is given by:

\[ \Delta T_B = K_B \cdot m \]

Where:
- \( K_B \) is the ebullioscopic constant (0.52 °C kg/mol for water)
- \( m \) is the molality of the solution

We can express molality (m) as:

\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]

Given that the mass of the solvent (water) is 500 g, we convert it to kg:

\[ \text{mass of solvent} = \frac{500}{1000} = 0.5 \text{ kg} \]

### Step 3: Rearranging to find molality
Now we can rearrange the boiling point elevation formula to find molality:

\[ m = \frac{\Delta T_B}{K_B} = \frac{0.104 °C}{0.52 °C \text{ kg/mol}} = 0.2 \text{ mol/kg} \]

### Step 4: Calculate moles of solute
Using the definition of molality:

\[ m = \frac{\text{moles of solute}}{0.5 \text{ kg}} \]

We can find the moles of solute:

\[ \text{moles of solute} = m \cdot \text{mass of solvent in kg} = 0.2 \text{ mol/kg} \cdot 0.5 \text{ kg} = 0.1 \text{ moles} \]

### Step 5: Calculate the molar mass of the solute
The molar mass (M) can be calculated using the formula:

\[ M = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{34.2 \text{ g}}{0.1 \text{ moles}} = 342 \text{ g/mol} \]

### Step 6: Calculate the depression in freezing point (ΔT_F)
The formula for freezing point depression is:

\[ \Delta T_F = K_F \cdot m \]

Where \( K_F \) for water is 1.87 °C kg/mol. We already calculated molality (m) as 0.2 mol/kg.

\[ \Delta T_F = 1.87 °C \text{ kg/mol} \cdot 0.2 \text{ mol/kg} = 0.374 °C \]

### Step 7: Calculate the freezing point of the solution
The freezing point of pure water (T_F₀) is 0 °C. Thus, the freezing point of the solution (T_F) is:

\[ T_F = T_F₀ - \Delta T_F = 0 °C - 0.374 °C = -0.374 °C \]

### Final Results
- Molar mass of the compound: **342 g/mol**
- Freezing point of the solution: **-0.374 °C**