Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving `1.0g` of polymer of molar mass `185,000` in `450mL` of water at `37^(@)C`.

Explanation

To calculate the osmotic pressure exerted by the solution, we will follow these steps:

### Step 1: Identify the given values
- Weight of solute (Wb) = 1.0 g
- Molar mass of solute (Mb) = 185,000 g/mol
- Volume of solution (V) = 450 mL = 0.450 L (since 1 L = 1000 mL)
- Temperature (T) = 37°C = 310 K (since K = °C + 273)

### Step 2: Calculate the number of moles of solute (N)
Using the formula:
\[ N = \frac{Wb}{Mb} \]
Substituting the values:
\[ N = \frac{1.0 \, \text{g}}{185,000 \, \text{g/mol}} = 5.405 \times 10^{-6} \, \text{mol} \]

### Step 3: Calculate the concentration (C) of the solution
The concentration \(C\) is given by:
\[ C = \frac{N}{V} \]
Substituting the values:
\[ C = \frac{5.405 \times 10^{-6} \, \text{mol}}{0.450 \, \text{L}} = 1.202 \times 10^{-5} \, \text{mol/L} \]

### Step 4: Use the osmotic pressure formula
The osmotic pressure (\(\pi\)) is given by:
\[ \pi = C \cdot R \cdot T \]
Where:
- \(R\) (the universal gas constant) = 8.314 J/(mol·K) = 8.314 × 10³ Pa·L/(mol·K)

Substituting the values:
\[ \pi = (1.202 \times 10^{-5} \, \text{mol/L}) \cdot (8.314 \times 10^3 \, \text{Pa·L/(mol·K)}) \cdot (310 \, \text{K}) \]

### Step 5: Calculate the osmotic pressure
Calculating:
\[ \pi = 1.202 \times 10^{-5} \cdot 8.314 \times 10^3 \cdot 310 \]
\[ \pi = 1.202 \times 10^{-5} \cdot 2577.34 \approx 30.96 \, \text{Pa} \]

### Final Answer
The osmotic pressure exerted by the solution is approximately **30.96 Pa**.

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