The osmotic pressure of a solution containing 9.2g of a substance (molar mass = 176) in 302 ml of solution was found to be 4.1 atmosphere at `15.5^(@)C`. Calculate the value of solution constant.

Explanation

To solve the problem of calculating the solution constant (R) using the given osmotic pressure, we will follow these steps:

### Step-by-Step Solution:

1. **Convert Temperature to Kelvin:**
The temperature is given as \( 15.5^\circ C \).
\[ T(K) = T(°C) + 273.15 = 15.5 + 273.15 = 288.65 \, K \]

2. **Identify Given Values:**
- Mass of solute (Wb) = 9.2 g
- Molar mass of solute (Mb) = 176 g/mol
- Volume of solution (V) = 302 mL = 0.302 L (since 1 L = 1000 mL)
- Osmotic pressure (π) = 4.1 atm

3. **Use the Osmotic Pressure Formula:**
The formula for osmotic pressure is given by:
\[ \pi = CRT \]
Rearranging this formula to solve for R gives:
\[ R = \frac{\pi}{C \cdot T} \]

4. **Calculate the Concentration (C):**
The concentration (C) can be calculated using the formula:
\[ C = \frac{Wb}{Mb \cdot V} \]
Substituting the values:
\[ C = \frac{9.2 \, g}{176 \, g/mol \cdot 0.302 \, L} = \frac{9.2}{53.152} \approx 0.173 \, mol/L \]

5. **Substitute Values into the R Formula:**
Now substitute π, C, and T into the rearranged formula for R:
\[ R = \frac{4.1 \, atm}{0.173 \, mol/L \cdot 288.65 \, K} \]

6. **Calculate R:**
\[ R = \frac{4.1}{0.173 \cdot 288.65} \approx \frac{4.1}{49.89} \approx 0.0821 \, L \cdot atm/(K \cdot mol) \]

### Final Result:
The value of the solution constant (R) is approximately:
\[ R \approx 0.0821 \, L \cdot atm/(K \cdot mol) \]

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