Calculate the osmotic pressure of a solutions containing 10 gram each of glucose `(C_(6)H_(12)O_(6))` and `(C_(12)H_(22)O_(11))` in `1000 cm^(3)` of the solution at `25^(@)C` .(`R =0.083L bar K^(-1) mol^(-1)`).

Explanation

To calculate the osmotic pressure of a solution containing 10 grams each of glucose (C₆H₁₂O₆) and sucrose (C₁₂H₂₂O₁₁) in 1000 cm³ of solution at 25°C, we will follow these steps:

### Step 1: Calculate the molar mass of glucose (C₆H₁₂O₆)
- Molar mass of glucose = (6 × 12) + (12 × 1) + (6 × 16)
- Molar mass of glucose = 72 + 12 + 96 = 180 g/mol

### Step 2: Calculate the molar mass of sucrose (C₁₂H₂₂O₁₁)
- Molar mass of sucrose = (12 × 12) + (22 × 1) + (11 × 16)
- Molar mass of sucrose = 144 + 22 + 176 = 342 g/mol

### Step 3: Calculate the number of moles of glucose
- Number of moles of glucose = mass (g) / molar mass (g/mol)
- Number of moles of glucose = 10 g / 180 g/mol = 0.0556 mol

### Step 4: Calculate the number of moles of sucrose
- Number of moles of sucrose = mass (g) / molar mass (g/mol)
- Number of moles of sucrose = 10 g / 342 g/mol = 0.0292 mol

### Step 5: Calculate the total number of moles in the solution
- Total number of moles = moles of glucose + moles of sucrose
- Total number of moles = 0.0556 mol + 0.0292 mol = 0.0848 mol

### Step 6: Convert the volume of the solution from cm³ to liters
- Volume of solution = 1000 cm³ = 1 L

### Step 7: Calculate the concentration (C) of the solution
- Concentration (C) = Total number of moles / Volume (L)
- Concentration (C) = 0.0848 mol / 1 L = 0.0848 mol/L

### Step 8: Convert the temperature from Celsius to Kelvin
- Temperature (T) = 25°C + 273 = 298 K

### Step 9: Use the osmotic pressure formula to calculate osmotic pressure (π)
- The formula for osmotic pressure is:
\[ \pi = C \cdot R \cdot T \]
- Where:
- C = concentration in mol/L = 0.0848 mol/L
- R = universal gas constant = 0.083 L bar K⁻¹ mol⁻¹
- T = temperature in K = 298 K

### Step 10: Substitute the values into the osmotic pressure formula
- \[ \pi = 0.0848 \, \text{mol/L} \times 0.083 \, \text{L bar K}^{-1} \text{mol}^{-1} \times 298 \, \text{K} \]
- \[ \pi = 2.10 \, \text{bar} \]

### Final Answer
The osmotic pressure of the solution is **2.10 bar**.