The mass of sodium acetate (CH₃COONa) required to prepare 250 mL of 0.35 M aqueous solution is ______ g. (Molar mass of CH₃COONa is 82.02 g mol⁻¹)

Sodium Acetate Molarity Numerical | Mole Concept | JEE Mains PYQ

Q. The mass of sodium acetate (CH₃COONa) required to prepare 250 mL of 0.35 M aqueous solution is ______ g. (Molar mass of CH₃COONa is 82.02 g mol⁻¹)

Step by Step Solution

Step 1: Convert volume into litre

250 mL = 0.25 L

Step 2: Calculate number of moles required

Moles = Molarity × Volume = 0.35 × 0.25 = 0.0875 mol

Step 3: Calculate mass of sodium acetate

Mass = Moles × Molar mass = 0.0875 × 82.02 ≈ 7.17 g

Step 4: Nearest integer value

Required mass ≈ 7 g

Final Answer: 7

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