Osmotic pressure of `30%` solution of glucose is `1.20` atm and that of `3.42%` solutin of cane sugar is `2.5` atm, The osmotic pressure of the mixture containing equal volumes of the two solutions will be

Options

A. `2.5` atm
B. `3.7` atm
C. `1.85` atm
D. `1.3` atm.

Answer

To find the osmotic pressure of a mixture containing equal volumes of a 30% glucose solution and a 3.42% cane sugar solution, we can follow these steps:

### Step 1: Understand the given data
- Osmotic pressure of the glucose solution (π₁) = 1.20 atm
- Osmotic pressure of the cane sugar solution (π₂) = 2.5 atm
- Both solutions are mixed in equal volumes.

### Step 2: Define the volumes
Let the volume of each solution be \( V \). Therefore, the total volume of the mixture will be:
\[ V_{\text{total}} = V + V = 2V \]

### Step 3: Use the formula for osmotic pressure
The osmotic pressure of the mixture (π_f) can be calculated using the formula:
\[ \pi_f \cdot V_{\text{total}} = \pi_1 \cdot V + \pi_2 \cdot V \]

### Step 4: Substitute the values into the equation
Substituting the known values into the equation:
\[ \pi_f \cdot (2V) = (1.20 \, \text{atm}) \cdot V + (2.5 \, \text{atm}) \cdot V \]

### Step 5: Simplify the equation
This simplifies to:
\[ \pi_f \cdot (2V) = (1.20 + 2.5) \cdot V \]
\[ \pi_f \cdot (2V) = 3.70 \cdot V \]

### Step 6: Solve for π_f
Now, divide both sides by \( 2V \):
\[ \pi_f = \frac{3.70 \cdot V}{2V} \]
\[ \pi_f = \frac{3.70}{2} \]
\[ \pi_f = 1.85 \, \text{atm} \]

### Conclusion
The osmotic pressure of the mixture is **1.85 atm**.

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