At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?

To find the concentration of the glucose solution at an osmotic pressure of 1.52 bars, we can use the formula for osmotic pressure, which is given by the equation:

Understanding Osmotic Pressure

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

\( \Pi = iCRT \)

Where:

• \( \Pi \) = osmotic pressure (in bars)
• i = van 't Hoff factor (which is 1 for glucose since it does not dissociate in solution)
• C = molarity of the solution (in moles per liter)
• R = ideal gas constant (0.0831 L·bar/K·mol)
• T = temperature (in Kelvin)

Calculating Molarity from Given Data

First, let's determine the molarity of the glucose solution when the osmotic pressure is 4.98 bars. We can rearrange the formula to solve for \( C \):

\( C = \frac{\Pi}{iRT} \)

Substituting the known values:

• \( \Pi = 4.98 \, \text{bar} \)
• \( i = 1 \)
• \( R = 0.0831 \, \text{L·bar/K·mol} \)
• \( T = 300 \, \text{K} \)

Now, plug in the values:

\( C = \frac{4.98}{1 \times 0.0831 \times 300} \)

Calculating this gives:

\( C = \frac{4.98}{24.93} \approx 0.199 \, \text{mol/L} \)

Finding the Concentration at 1.52 Bars

Now, we can use the same formula to find the concentration when the osmotic pressure is 1.52 bars:

\( C = \frac{1.52}{1 \times 0.0831 \times 300} \)

Calculating this gives:

\( C = \frac{1.52}{24.93} \approx 0.061 \, \text{mol/L} \)

Final Result

Thus, the concentration of the glucose solution at an osmotic pressure of 1.52 bars is approximately 0.061 mol/L.

This calculation illustrates how osmotic pressure is directly proportional to the concentration of solute in a solution, allowing us to determine the concentration based on changes in pressure.