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?

• A. 11 g L⁻¹
• B. 22 g L⁻¹
• C. 36 g L⁻¹
• D. 42 g L⁻¹

To find the concentration of glucose in the solution with an osmotic pressure of 1.52 bars, we can use the formula for osmotic pressure:

Osmotic Pressure Formula

The osmotic pressure (\( \Pi \)) is given by the equation:

\( \Pi = iCRT \)

Where:

• \( i \) = van 't Hoff factor (for glucose, \( i = 1 \))
• C = concentration in mol/L
• R = ideal gas constant (0.0831 L·bar/(K·mol))
• T = temperature in Kelvin

Calculating Concentration

We know the osmotic pressure at 300 K is 4.98 bar for 36 g of glucose. First, we calculate the molarity of the original solution:

The molar mass of glucose (C₆H₁₂O₆) is approximately 180 g/mol. Thus, the number of moles in 36 g is:

Number of moles = \( \frac{36 \text{ g}}{180 \text{ g/mol}} = 0.2 \text{ mol} \)

The concentration of the original solution is:

C = \( \frac{0.2 \text{ mol}}{1 \text{ L}} = 0.2 \text{ mol/L} \)

Finding New Concentration

Now, we can find the new concentration using the osmotic pressure of 1.52 bars:

Rearranging the osmotic pressure formula gives:

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

Substituting the values:

C = \( \frac{1.52 \text{ bar}}{(1)(0.0831 \text{ L·bar/(K·mol)})(300 \text{ K})} \)

Calculating this gives:

C ≈ 0.061 mol/L

Converting to Grams per Liter

Now, convert the concentration from mol/L to g/L:

Concentration in g/L = \( 0.061 \text{ mol/L} \times 180 \text{ g/mol} ≈ 11 \text{ g/L} \)

Final Answer

The concentration of the solution with an osmotic pressure of 1.52 bars is 11 g L⁻¹, which corresponds to option A.