What is the osmotic pressure of a solution of `4.48g` of a substance of molecular weight `286` in 100`cm^(3)` water at `298K`?

Explanation

To find the osmotic pressure of the solution, we can follow these steps:

### Step 1: Calculate the number of moles of the solute.
The number of moles (n) can be calculated using the formula:
\[ n = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)}} \]
Given:
- Mass of solute = 4.48 g
- Molar mass = 286 g/mol

Substituting the values:
\[ n = \frac{4.48 \, \text{g}}{286 \, \text{g/mol}} \approx 0.0157 \, \text{mol} \]

### Step 2: Calculate the concentration of the solution.
Concentration (C) in mol/L can be calculated using the formula:
\[ C = \frac{n}{V} \]
Where:
- \(n\) = number of moles
- \(V\) = volume of solution in liters

Given:
- Volume = 100 cm³ = 0.1 L

Substituting the values:
\[ C = \frac{0.0157 \, \text{mol}}{0.1 \, \text{L}} = 0.157 \, \text{mol/L} \]

### Step 3: Use the osmotic pressure formula.
The osmotic pressure (\(\Pi\)) can be calculated using the formula:
\[ \Pi = C \cdot R \cdot T \]
Where:
- \(C\) = concentration in mol/L
- \(R\) = gas constant = 8.31 L·kPa/(K·mol) or 0.0821 L·atm/(K·mol)
- \(T\) = temperature in Kelvin

Given:
- \(C = 0.157 \, \text{mol/L}\)
- \(R = 0.0821 \, \text{L·atm/(K·mol)}\)
- \(T = 298 \, \text{K}\)

Substituting the values:
\[ \Pi = 0.157 \, \text{mol/L} \cdot 0.0821 \, \text{L·atm/(K·mol)} \cdot 298 \, \text{K} \]
Calculating this gives:
\[ \Pi \approx 3.87 \, \text{atm} \]

### Final Answer:
The osmotic pressure of the solution is approximately **3.87 atm**.

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