Calculate the osmotic pressure at `25^(@)C` of a solution containing 1g of glucose and 1g of sucrose in 1 litre of solution.
If it were not known that the solute was a mixture of glucose and sucrose, what would be the molecular weight of solute corresponding to the calculated osmotic pressure ?

Explanation

To solve the problem, we will calculate the osmotic pressure of a solution containing 1g of glucose and 1g of sucrose in 1 liter of solution at 25°C. We will also determine the molecular weight of the solute if it were not known to be a mixture of glucose and sucrose.

### Step 1: Calculate the number of moles of glucose and sucrose.

1. **Molecular weight of glucose (C6H12O6)** = 180 g/mol
2. **Molecular weight of sucrose (C12H22O11)** = 342 g/mol

Using the formula for the number of moles:
\[ n = \frac{\text{mass (g)}}{\text{molecular weight (g/mol)}} \]

- For glucose:
\[ n_1 = \frac{1 \text{ g}}{180 \text{ g/mol}} = \frac{1}{180} \text{ mol} \]

- For sucrose:
\[ n_2 = \frac{1 \text{ g}}{342 \text{ g/mol}} = \frac{1}{342} \text{ mol} \]

### Step 2: Calculate the total number of moles in the solution.

\[ n_{\text{total}} = n_1 + n_2 = \frac{1}{180} + \frac{1}{342} \]

To calculate this, we find a common denominator:
\[ n_{\text{total}} = \frac{342 + 180}{180 \times 342} = \frac{522}{61560} \text{ mol} \]

### Step 3: Use the osmotic pressure formula.

The formula for osmotic pressure (\(\Pi\)) is given by:
\[ \Pi = \frac{nRT}{V} \]
Where:
- \(R\) = 0.0821 L·atm/(K·mol)
- \(T\) = 25°C = 298 K
- \(V\) = 1 L

Substituting the values:
\[ \Pi = \frac{n_{\text{total}} \cdot R \cdot T}{V} = \frac{\left(\frac{522}{61560}\right) \cdot 0.0821 \cdot 298}{1} \]

Calculating this gives:
\[ \Pi \approx 0.2078 \text{ atm} \]

### Step 4: Calculate the molecular weight of the solute if it were unknown.

Using the osmotic pressure formula rearranged:
\[ \Pi V = nRT \]
We can find \(n\) (number of moles) as:
\[ n = \frac{\Pi V}{RT} \]

Substituting the known values:
\[ n = \frac{0.2078 \cdot 1}{0.0821 \cdot 298} \]
Calculating this gives:
\[ n \approx 0.008 \text{ mol} \]

### Step 5: Calculate the molecular weight of the solute.

The molecular weight (M) can be calculated using the formula:
\[ M = \frac{\text{mass (g)}}{n} \]

The total mass of the solute is:
\[ \text{mass} = 1 \text{ g (glucose)} + 1 \text{ g (sucrose)} = 2 \text{ g} \]

Thus,
\[ M = \frac{2 \text{ g}}{0.008 \text{ mol}} = 250 \text{ g/mol} \]

### Final Answers:
1. The osmotic pressure of the solution is approximately **0.2078 atm**.
2. The molecular weight of the solute, if unknown, would be approximately **250 g/mol**.