The osmatic pressure of a non-volatile solute 'X' in benezene at `25^@C` is `20.66 N m^(-2)` . If the concentration of solution is `2 gL^(-1)` , what molecular weight of X ?

Answer

To find the molecular weight of the non-volatile solute 'X' in benzene, we can use the formula for osmotic pressure:

\[ \pi = cRT \]

Where:
- \(\pi\) = osmotic pressure (in N/m²)
- \(c\) = concentration (in moles per liter)
- \(R\) = universal gas constant (8.314 J/(mol·K))
- \(T\) = temperature (in Kelvin)

### Step-by-Step Solution:

1. **Identify Given Values**:
- Osmotic pressure (\(\pi\)) = 20.66 N/m²
- Concentration (c) = 2 g/L
- Temperature (T) = 25°C

2. **Convert Temperature to Kelvin**:
\[ T = 25 + 273 = 298 \text{ K} \]

3. **Express Concentration in Terms of Molarity**:
The concentration in terms of molarity (moles per liter) can be expressed as:
\[ c = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)} \times \text{volume (L)}} \]
Given that the concentration is 2 g/L, we can express it as:
\[ c = \frac{2 \text{ g}}{M \text{ g/mol}} = \frac{2}{M} \text{ mol/L} \]
where \(M\) is the molar mass of solute 'X'.

4. **Substitute Values into the Osmotic Pressure Formula**:
Substitute \(c\) into the osmotic pressure equation:
\[ \pi = \left(\frac{2}{M}\right)RT \]
Rearranging gives:
\[ M = \frac{2RT}{\pi} \]

5. **Substitute Known Values**:
Now, substitute \(R = 8.314 \text{ J/(mol·K)}\), \(T = 298 \text{ K}\), and \(\pi = 20.66 \text{ N/m²}\):
\[ M = \frac{2 \times 8.314 \times 298}{20.66} \]

6. **Calculate the Molar Mass**:
Performing the calculation:
\[ M = \frac{2 \times 8.314 \times 298}{20.66} \approx 239.8 \text{ g/mol} \]

7. **Final Result**:
The molecular weight of solute 'X' is approximately:
\[ M \approx 239.8 \text{ g/mol} \]

### Summary:
The molecular weight of the non-volatile solute 'X' is **239.8 g/mol**.