Question

The osmotic pressure of a dilute solution of a compound XY in water is four times that of a solution of 0.01 M BaCl₂ in water. Assuming complete dissociation of the given ionic compounds in water, the concentration of XY (in mol/L) in solution is:

A. 4×10⁻²
B. 6×10⁻²
C. 4×10⁻⁴
D. 16×10⁻⁴

Answer 1

The concentration of the compound XY, which has an osmotic pressure four times that of a 0.01 M BaCl2 solution and assuming complete dissociation into two ions, is 0.06 M.

Step-by-step explanation:

The student's question addresses the comparison of osmotic pressure between two solutions: one containing the compound XY and the other containing BaCl2. Since osmotic pressure depends on the concentration of solute particles, we can deduce the concentration of XY by understanding how its osmotic pressure compares to the osmotic pressure of a known concentration of BaCl2.

Given that the osmotic pressure of the XY solution is four times that of a 0.01 M BaCl2 solution, and BaCl2 completely dissociates to form 3 solute particles in solution (1 Ba2+ and 2 Cl-), we can calculate the effective concentration of solute particles of the BaCl2 solution to be 0.01 M x 3 = 0.03 M. Since the osmotic pressure is directly proportional to the solute particle concentration, the concentration of XY that would produce four times the osmotic pressure of the BaCl2 solution is 0.03 M x 4 = 0.12 M. However, we need to account for the dissociation of XY into its constituent ions. If XY dissociates completely into X+ and Y-, each mole of XY would produce two moles of solute particles, thus the required concentration of undissociated XY would be 0.12 M / 2 = 0.06 M.