A living cell contains a solution which is isotonic with 0.2 M glucose solution. What osmotic pressure develops when the cell is placed in 0.05 M `BaCl_(2)`solution at 300 K ?
(a)1.23 atm
(b)3.69 atm
(c)6.15 atm
(d)None of these

Options

A. 1.23 atm
B. 3.69 atm
C. 6.15 atm
D. None of these

Answer

To solve the problem, we need to calculate the osmotic pressure developed when a living cell containing a solution isotonic with 0.2 M glucose is placed in a 0.05 M BaCl₂ solution at 300 K. We will use the formula for osmotic pressure, which is given by:

\[ \Pi = iCRT \]

Where:
- \(\Pi\) = osmotic pressure
- \(i\) = van 't Hoff factor (number of particles the solute dissociates into)
- \(C\) = molarity of the solution
- \(R\) = ideal gas constant (0.0821 L·atm/(K·mol))
- \(T\) = temperature in Kelvin

### Step 1: Calculate the osmotic pressure of the glucose solution
For glucose:
- Concentration \(C = 0.2 \, \text{M}\)
- van 't Hoff factor \(i = 1\) (since glucose does not dissociate)

Using the formula:
\[ \Pi_{\text{glucose}} = i \cdot C \cdot R \cdot T \]
\[ \Pi_{\text{glucose}} = 1 \cdot 0.2 \cdot 0.0821 \cdot 300 \]
\[ \Pi_{\text{glucose}} = 4.926 \, \text{atm} \]

### Step 2: Calculate the osmotic pressure of the BaCl₂ solution
For BaCl₂:
- Concentration \(C = 0.05 \, \text{M}\)
- van 't Hoff factor \(i = 3\) (BaCl₂ dissociates into 1 Ba²⁺ and 2 Cl⁻ ions)

Using the formula:
\[ \Pi_{\text{BaCl2}} = i \cdot C \cdot R \cdot T \]
\[ \Pi_{\text{BaCl2}} = 3 \cdot 0.05 \cdot 0.0821 \cdot 300 \]
\[ \Pi_{\text{BaCl2}} = 3 \cdot 0.05 \cdot 24.63 \]
\[ \Pi_{\text{BaCl2}} = 3.69 \, \text{atm} \]

### Step 3: Total osmotic pressure
Now, we add the osmotic pressures from both solutions:
\[ \Pi_{\text{total}} = \Pi_{\text{glucose}} + \Pi_{\text{BaCl2}} \]
\[ \Pi_{\text{total}} = 4.926 + 3.69 = 8.616 \, \text{atm} \]

### Conclusion
Since the calculated osmotic pressure (8.616 atm) does not match any of the provided options (1.23 atm, 3.69 atm, 6.15 atm), the answer is:
**(d) None of these.**