A solution of an organic compound is prepared by dissolving 68.4 g in 1000 g of water.
Calculate the molecular mass of the compound and osmotic pressure of the solution
at 293 K when elevation of b.pt is 0.104 and `K_(b)` for water is `"0.52 K mol"^(-1)`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the molality of the solution
The formula for elevation in boiling point is given by:

\[ \Delta T_b = K_b \times m \]

Where:
- \(\Delta T_b\) = elevation in boiling point
- \(K_b\) = ebullioscopic constant
- \(m\) = molality of the solution

Given:
- \(\Delta T_b = 0.104 \, \text{K}\)
- \(K_b = 0.52 \, \text{K kg mol}^{-1}\)

Rearranging the formula to find molality (\(m\)):

\[ m = \frac{\Delta T_b}{K_b} \]

Substituting the values:

\[ m = \frac{0.104}{0.52} = 0.2 \, \text{mol/kg} \]

### Step 2: Calculate the number of moles of solute
Molality is defined as the number of moles of solute per kilogram of solvent. We have 1000 g of water, which is 1 kg.

Using the formula:

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

Where \(n\) is the number of moles of solute. Rearranging gives:

\[ n = m \times \text{mass of solvent in kg} \]

Substituting the values:

\[ n = 0.2 \, \text{mol/kg} \times 1 \, \text{kg} = 0.2 \, \text{mol} \]

### Step 3: Calculate the molecular mass of the compound
The molecular mass (M) can be calculated using the formula:

\[ M = \frac{\text{mass of solute (g)}}{\text{number of moles (mol)}} \]

Given that the mass of the solute is 68.4 g:

\[ M = \frac{68.4 \, \text{g}}{0.2 \, \text{mol}} = 342 \, \text{g/mol} \]

### Step 4: Calculate the osmotic pressure of the solution
The formula for osmotic pressure (\(\Pi\)) is given by:

\[ \Pi = nCRT \]

Where:
- \(n\) = number of moles of solute
- \(C\) = concentration in mol/L (which is equal to \(n/V\), where \(V\) is the volume in liters)
- \(R\) = universal gas constant = 0.0821 L·atm/(K·mol)
- \(T\) = temperature in Kelvin

Since we have 68.4 g of solute in 1000 g of water, we can assume the volume of the solution is approximately 1 L (for dilute solutions).

Now, substituting the values:

\[ \Pi = nRT = 0.2 \, \text{mol} \times 0.0821 \, \text{L·atm/(K·mol)} \times 293 \, \text{K} \]

Calculating:

\[ \Pi = 0.2 \times 0.0821 \times 293 \approx 4.83 \, \text{atm} \]

### Final Answers
- Molecular mass of the compound = 342 g/mol
- Osmotic pressure of the solution = 4.83 atm