Molarity to Molality Conversion | CuSO4 Numerical | JEE Mains PYQ
Q. Molarity (M) of an aqueous solution containing x g of anhydrous CuSO₄ in 500 mL solution at 32°C is 2 × 10⁻¹ M. Its molality will be ________ × 10⁻³ m. (nearest integer) [Given density of the solution = 1.25 g/mL]
Step by Step Solution
Step 1: Calculate moles of CuSO₄

Molarity = 0.2 M Volume = 500 mL = 0.5 L Moles of CuSO₄ = 0.2 × 0.5 = 0.1 mol

Step 2: Calculate mass of solution

Density = 1.25 g/mL Mass of solution = 1.25 × 500 = 625 g

Step 3: Calculate mass of solute

Molar mass of anhydrous CuSO₄ = 160 g/mol Mass of solute = 0.1 × 160 = 16 g

Step 4: Calculate mass of solvent

Mass of solvent = 625 − 16 = 609 g = 0.609 kg

Step 5: Calculate molality

Molality = moles of solute / kg of solvent = 0.1 / 0.609 ≈ 0.164 m = 164 × 10⁻³ m

Final Answer: 164

Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.

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