Question

At 300 K, 36 g of glucose present in a liter 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. 0.5 M
b. 1.5 M
c. 2.0 M
d. 3.0 M

Answer 1

The concentration of the glucose solution with an osmotic pressure of 1.52 bars at 300 K is calculated to be 0.061 M using the osmotic pressure formula and the given osmotic pressure at the same temperature.

Step-by-step explanation:

To find the concentration of the glucose solution when the osmotic pressure is 1.52 bars, we make use of the fact that osmotic pressure (Π) is directly proportional to the molar concentration (C) of the solution when the temperature (T) remains constant. The formula for osmotic pressure is Π = MRT, where M is the molarity of the solution, R is the ideal gas constant, and T is the temperature.

Given that at 300 K, 36 g of glucose (molecular weight = 180 g/mol) has an osmotic pressure of 4.98 bar, we can determine the molarity by dividing the mass by the molar mass to obtain moles of glucose, and then by volume to get the molarity, which is 0.2 M. If the osmotic pressure is inversely proportional to it, we can set up a ratio using the two given pressures (4.98 bars and 1.52 bars).

The new concentration (C2) can be calculated using the formula:

Π1/Π2 = C1/C2

Plugging in the values we have:

4.98 bar / 1.52 bar = 0.2 M / C2

Thus, C2 = (0.2 M * 1.52 bar) / 4.98 bar = 0.061 M.