Match List-I with List-II: List-I (Solution preparation) List-II (Molarity) (A) 4.0 ~g ~NaOH in 250 ~mL (I) 0.1 M (B) 34.2 ~g Sucrose in 1 ~L (II) 0.2 M (C) 3.15 ~g HNO ₃ in 250 ~mL (III) 0.4 M (D) 9.0 ~g Glucose in 100 ~mL (IV) 0.5 M (Molar masses in g mol ⁻¹ : NaOH =40 , Sucrose =342 , HNO ₃=63 , Glucose =180 )

Step-by-step solution

Molarity ( M ) = Mass of solute Molar mass Volume in L (A) For NaOH : M = 4.0 40 0.25 = 0.1 0.25 = 0.4 M (Matches III) (B) For Sucrose: M = 34.2 342 1 = 0.1 M (Matches I) (C) For HNO ₃ : M = 3.15 63 0.25 = 0.05 0.25 = 0.2 M (Matches II) (D) For Glucose: M = 9.0 180 0.1 = 0.05 0.1 = 0.5 M (Matches IV) Therefore, the correct matching is (A)-(III), (B)-(I), (C)-(II), (D)-(IV). Answer: (A)-(III), (B)-(I), (C)-(II), (D)-(IV)