How many `gm` of glucose must be present in `0.45` litre of a solution for its osmotic pressure to be same as that of solution of `8.89 gm` glucose per litre?

Explanation

To solve the problem of how many grams of glucose must be present in 0.45 liters of a solution for its osmotic pressure to be the same as that of a solution containing 8.89 grams of glucose per liter, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Osmotic Pressure Formula**:
The osmotic pressure (π) of a solution is given by the formula:
\[ \pi = CRT \]
where \(C\) is the concentration in moles per liter, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin. Since \(R\) and \(T\) are constant for both solutions, we can equate the concentrations.

2. **Calculate Concentration of the Second Solution**:
For the second solution, which has 8.89 grams of glucose per liter:
- The molar mass of glucose (C₆H₁₂O₆) is approximately 180 g/mol.
- The number of moles of glucose in 8.89 grams is:
\[ \text{Moles} = \frac{8.89 \text{ g}}{180 \text{ g/mol}} \approx 0.0494 \text{ moles} \]
- Therefore, the concentration \(C_2\) of the second solution is:
\[ C_2 = \frac{0.0494 \text{ moles}}{1 \text{ L}} = 0.0494 \text{ M} \]

3. **Set Up the Equation for the First Solution**:
Let \(W_1\) be the mass of glucose in grams that we need to find for the first solution (0.45 L).
- The concentration \(C_1\) of the first solution can be expressed as:
\[ C_1 = \frac{W_1 / 180}{0.45} \]
where \(W_1 / 180\) is the number of moles of glucose in \(W_1\) grams.

4. **Equate the Concentrations**:
Since the osmotic pressures are equal, we can set the concentrations equal to each other:
\[ C_1 = C_2 \]
\[ \frac{W_1 / 180}{0.45} = 0.0494 \]

5. **Solve for \(W_1\)**:
Rearranging the equation gives:
\[ W_1 = 0.0494 \times 0.45 \times 180 \]
Now, calculate \(W_1\):
\[ W_1 = 0.0494 \times 0.45 \times 180 \approx 4.0 \text{ g} \]

### Final Answer:
Thus, approximately **4 grams** of glucose must be present in 0.45 liters of the solution.