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?

The given problem is a typical application of the formula for osmotic pressure:

π = n/V * R * T

where: π is the osmotic pressure, n is the number of moles of the solute, V is the volume of the solution in liters, R is the ideal gas constant, and T is the temperature in Kelvin.

Step 1: Calculate the number of moles of glucose in the initial solution.

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

n = mass/molar mass = 36 g / 180 g/mol = 0.2 mol

Step 2: Substitute the known values into the osmotic pressure formula to find the value of R.

4.98 bar = 0.2 mol/1 L * R * 300 K

Solving for R gives R = 4.98 bar / (0.2 mol/1 L * 300 K) = 0.083 bar L / (mol K)

Step 3: Use the value of R to find the number of moles of glucose in the new solution.

1.52 bar = n/1 L * 0.083 bar L / (mol K) * 300 K

Solving for n gives n = 1.52 bar / (0.083 bar L / (mol K) * 300 K) = 0.061 mol

Step 4: The concentration of the new solution is the number of moles per liter, which is 0.061 mol/L.