`10.1g` of a volatile liquid occupies a volume of 4 litres when vaporised at `100^(@)C` and 70 cm pressure. What would be the osmotic pressure of a `2%` (grams per 100 "cc") solution of this substance at `0^(@)C`?

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Convert Pressure from cm Hg to atm
Given pressure \( P = 70 \, \text{cm Hg} \).

To convert cm Hg to atm:
\[ P = \frac{70 \, \text{cm Hg}}{76 \, \text{cm Hg/atm}} = 0.9211 \, \text{atm} \]

### Step 2: Convert Temperature from Celsius to Kelvin
Given temperature \( T = 100^\circ C \).

To convert Celsius to Kelvin:
\[ T = 100 + 273 = 373 \, \text{K} \]

### Step 3: Calculate the Molar Mass of the Volatile Liquid
Using the Ideal Gas Law \( PV = nRT \), we can express the number of moles \( n \) as:
\[ n = \frac{PV}{RT} \]

Substituting the values:
\[ n = \frac{(0.9211 \, \text{atm}) \times (4 \, \text{L})}{(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}) \times (373 \, \text{K})} \]

Calculating \( n \):
\[ n = \frac{3.6844}{30.6583} \approx 0.120 \, \text{mol} \]

Now, we can find the molar mass \( M \) of the volatile liquid using the formula:
\[ M = \frac{\text{mass}}{n} = \frac{10.1 \, \text{g}}{0.120 \, \text{mol}} \approx 84.17 \, \text{g/mol} \]

### Step 4: Calculate the Molarity of the 2% Solution
Given a 2% solution means 2 grams of solute per 100 mL of solution.

To find the molarity \( C \):
\[ C = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)} \times \text{volume (L)}} \]
\[ C = \frac{2 \, \text{g}}{84.17 \, \text{g/mol} \times 0.1 \, \text{L}} = \frac{2}{8.417} \approx 0.237 \, \text{mol/L} \]

### Step 5: Calculate the Osmotic Pressure
Using the formula for osmotic pressure \( \Pi \):
\[ \Pi = C \cdot R \cdot T \]
Where:
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 0^\circ C = 273 \, \text{K} \)

Substituting the values:
\[ \Pi = 0.237 \, \text{mol/L} \cdot 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \cdot 273 \, \text{K} \]
\[ \Pi \approx 5.336 \, \text{atm} \]

### Final Answer
The osmotic pressure of the 2% solution at \( 0^\circ C \) is approximately \( 5.336 \, \text{atm} \).

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