The water vapour pressure at `293K` is `2338` Pa and the vapour pressure of an aqueous solution is `2295.8Pa`. Determine its osmotic pressure at 313 K if the solution density at this temperature is `1010 kg//m^(3)`. The molecular weight of the solute is 60

Answer

To determine the osmotic pressure of the solution at 313 K, we will follow these steps:

### Step 1: Calculate the mole fraction of the solute

The mole fraction of the solute (\(X_{solute}\)) can be calculated using the formula:

\[ X_{solute} = \frac{P_0 - P_{solution}}{P_0} \]

Where:
- \(P_0\) = vapor pressure of pure solvent (water) = 2338 Pa
- \(P_{solution}\) = vapor pressure of the solution = 2295.8 Pa

Substituting the values:

\[ X_{solute} = \frac{2338 - 2295.8}{2338} = \frac{42.2}{2338} \approx 0.018 \]

### Step 2: Calculate the mole fraction of the solvent

The mole fraction of the solvent (\(X_{solvent}\)) can be calculated as:

\[ X_{solvent} = 1 - X_{solute} \approx 1 - 0.018 = 0.982 \]

### Step 3: Calculate the number of moles of solute

To find the number of moles of solute, we need to use the density of the solution and the molecular weight of the solute.

1. Calculate the mass of 1 m³ of the solution:
\[ \text{Mass of solution} = \text{Density} \times \text{Volume} = 1010 \, \text{kg/m}^3 \times 1 \, \text{m}^3 = 1010 \, \text{kg} \]

2. Assume the mass of the solute is \(m\) kg. The mass of the solvent is then:
\[ \text{Mass of solvent} = 1010 - m \]

3. The number of moles of solute (\(n_{solute}\)) can be calculated using the molecular weight:
\[ n_{solute} = \frac{m}{M} = \frac{m}{60} \]

### Step 4: Calculate the osmotic pressure

The osmotic pressure (\(\Pi\)) can be calculated using the formula:

\[ \Pi = n_{solute} \cdot R \cdot T \]

Where:
- \(R\) = universal gas constant = 8.314 J/(mol·K)
- \(T\) = temperature in Kelvin = 313 K

Substituting \(n_{solute}\):

\[ \Pi = \left(\frac{m}{60}\right) \cdot 8.314 \cdot 313 \]

To find \(m\), we can use the mole fraction calculated earlier. The total number of moles in the solution can be expressed as:

\[ \text{Total moles} = \frac{n_{solute}}{X_{solute}} = \frac{n_{solute}}{0.018} \]

Assuming the mass of the solute is negligible compared to the solvent, we can approximate the total moles to be primarily from the solvent.

### Final Calculation

To find the osmotic pressure, we can express it in terms of the density and mole fraction:

\[ \Pi \approx \frac{X_{solute} \cdot \text{Density} \cdot R \cdot T}{M} \]

Substituting the values:

\[ \Pi \approx \frac{0.018 \cdot 1010 \cdot 8.314 \cdot 313}{60} \]

Calculating this gives:

\[ \Pi \approx \frac{0.018 \cdot 1010 \cdot 8.314 \cdot 313}{60} \approx 8.6 \, \text{Pa} \]

### Summary

The osmotic pressure of the solution at 313 K is approximately \(8.6 \, \text{Pa}\).

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