NEETChemistrySolutions
Match List-I with List-II, assuming complete dissociation of the electrolytes in List-I. List-I (Electrolyte solution) List-II (Isotonic glucose solution) (A) 0.1 M KCl (I) 0.4 M (B) 0.1 M CaCl ₂ (II) 0.5 M (C) 0.1 M FeCl ₃ (III) 0.2 M (D) 0.1 M Al ₂( SO ₄)₃ (IV) 0.3 M Choose the correct answer from the options given below:
Options
- A(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
- B(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
- C(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
- D(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
Correct answer
B. (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
Step-by-step solution
For isotonic solutions at a given temperature, the effective molar concentration of particles ( i C ) must be equal. For glucose (a non-electrolyte), the van't Hoff factor i = 1 . Thus, its effective concentration is equal to its molarity. For the strong electrolytes in List-I (assuming complete dissociation): (A) KCl K ^+ + Cl ^- ( i = 2 ). Effective concentration = 2 0.1 = 0.2 M . This matches with (III). (B) CaCl ₂ Ca ²⁺ + 2 Cl ^- ( i = 3 ). Effective concentration = 3 0.1 = 0.3 M . This matches with (IV). (C) F