Match List-I with List-II: List-I (Solute in water) List-II (Molality) A. 6 g Urea in 500 g water I. 1.0 m B. 18 g Glucose in 250 g water II. 0.05 m C. 34.2 g Sucrose in 100 g water III. 0.2 m D. 4 g NaOH in 2000 g water IV. 0.4 m Choose the correct answer from the options given below:

Step-by-step solution

Molality is defined as the number of moles of solute per kilogram of solvent: m = n_ solute W_ solvent ( in kg ) A. Urea ( NH ₂ CONH ₂ , molar mass = 60 g mol ⁻¹ ): Moles = 6 60 = 0.1 mol Molality = 0.1 mol 0.5 kg = 0.2 m (Matches III) B. Glucose ( C ₆ H ₁₂ O ₆ , molar mass = 180 g mol ⁻¹ ): Moles = 18 180 = 0.1 mol Molality = 0.1 mol 0.25 kg = 0.4 m (Matches IV) C. Sucrose ( C ₁₂ H ₂₂ O ₁₁ , molar mass = 342 g mol ⁻¹ ): Moles = 34.2 342 = 0.1 mol Molality = 0.1 mol 0.1 kg = 1.0 m (Matches I) D. NaOH (molar mass =