One litre aqueous solution of sucrose (molar mass = `"342 g mol"^(-1)`) weighing 1015 g is found to record an osmotic pressure of 4.82 atm at 292 K. What is the molality of the sucrose solution?
(R = 0.0821 atm `"mol"^(-1)"K"^(-1)`).

Explanation

To find the molality of the sucrose solution, we will follow these steps:

### Step 1: Calculate the concentration (C) using osmotic pressure formula
The formula for osmotic pressure (π) is given by:
\[ \pi = C \cdot R \cdot T \]
Where:
- π = osmotic pressure (4.82 atm)
- C = molarity (moles of solute per liter of solution)
- R = ideal gas constant (0.0821 atm L mol\(^{-1}\) K\(^{-1}\))
- T = temperature in Kelvin (292 K)

Rearranging the formula to find C:
\[ C = \frac{\pi}{R \cdot T} \]

Substituting the values:
\[ C = \frac{4.82 \, \text{atm}}{0.0821 \, \text{atm L mol}^{-1} \text{K}^{-1} \cdot 292 \, \text{K}} \]
Calculating:
\[ C \approx 0.2 \, \text{mol/L} \]

### Step 2: Find the number of moles of sucrose
Since the volume of the solution is 1 L, the number of moles (n) can be calculated as:
\[ n = C \cdot V = 0.2 \, \text{mol/L} \cdot 1 \, \text{L} = 0.2 \, \text{mol} \]

### Step 3: Calculate the mass of sucrose
Using the molar mass of sucrose (342 g/mol), we can find the mass of sucrose (w):
\[ w = n \cdot \text{molar mass} = 0.2 \, \text{mol} \cdot 342 \, \text{g/mol} = 68.4 \, \text{g} \]

### Step 4: Calculate the mass of the solvent
The total mass of the solution is given as 1015 g. The mass of the solvent (water) can be calculated as:
\[ \text{mass of solvent} = \text{mass of solution} - \text{mass of solute} = 1015 \, \text{g} - 68.4 \, \text{g} = 946.6 \, \text{g} \]

### Step 5: Convert the mass of solvent to kg
To calculate molality, we need the mass of the solvent in kilograms:
\[ \text{mass of solvent in kg} = \frac{946.6 \, \text{g}}{1000} = 0.9466 \, \text{kg} \]

### Step 6: Calculate the molality (m)
Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ m = \frac{n}{\text{mass of solvent in kg}} = \frac{0.2 \, \text{mol}}{0.9466 \, \text{kg}} \approx 0.2112 \, \text{mol/kg} \]

### Final Answer
The molality of the sucrose solution is approximately **0.2112 mol/kg**.

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