(B) < (A) < (D) < (C)
(C) < (B) < (A) < (D)
The osmotic pressure (\( \pi \)) of a solution can be determined using the formula:
\( \pi = iCRT \)
Where:
Since the temperature and gas constant are the same for all cases, the osmotic pressure depends on the molarity:
Arranging these molarities in increasing order gives: \( C_B < C_A < C_D < C_C \). Therefore, the solution with the lowest molarity will generate the lowest osmotic pressure, and the highest molarity will generate the highest osmotic pressure. Thus, in increasing order of osmotic pressure: (B) < (A) < (D) < (C).
The correct arrangement is: (B) < (A) < (D) < (C).
To determine the osmotic pressure generated by each solution, we use the formula for osmotic pressure, π: π = iCRT, where i is the van 't Hoff factor, C is the concentration of solute in mol/L, R is the ideal gas constant (0.0821 L atm K-1 mol-1), and T is the temperature in Kelvin (298 K). Since the cell wall is permeable to water and not to solute molecules, i = 1.0 for all solutions as there is no dissociation.
Let's calculate C for each case:
(A) 0.5 moles in 1 L: C = 0.5 M
(B) 0.25 moles in 1 L: C = 0.25 M
(C) 0.1 moles in 0.01 L: C = 10 M
(D) 0.2 moles in 0.05 L: C = 4 M
Substituting for each:
(A) π = 1 × 0.5 × 0.0821 × 298 = 12.239 atm
(B) π = 1 × 0.25 × 0.0821 × 298 = 6.119 atm
(C) π = 1 × 10 × 0.0821 × 298 = 244.58 atm
(D) π = 1 × 4 × 0.0821 × 298 = 97.912 atm
Now, arrange the osmotic pressures in increasing order: (B) (6.119 atm) < (A) (12.239 atm) < (D) (97.912 atm) < (C) (244.58 atm). Thus, the increasing order of osmotic pressure generation is: (B) < (A) < (D) < (C).
Hence, the correct answer is: (B) < (A) < (D) < (C)
The osmotic pressure of seawater is 1.05 atm. Four experiments were carried out as shown in the table. In which of the following experiments, pure water can be obtained in part-II of the vessel?