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NTA Abhyas JEE Main2020ChemistrySome Basic Concepts of ChemistryPractice

Calculate normality, molarity and molality of the solution containing 22% (w/w) salt A l 2 ( S O 4 ) 3 d = 1.253 g m / m l by weight is

Options

  1. A4.83 N , 8.25 M , 8.25 m
  2. B48.3 N , 0.825 M , 0.825 m
  3. C4.83 N , 4.83 M , 4.83 m
  4. D4.83 N , 0.805 M , 0.825 m

Correct answer

D. 4.83 N , 0.805 M , 0.825 m

Step-by-step solution

Normality = w / w % × 10 × d E 2 = 22 × 10 × 1.253 342 6 = 4.83 N Molarity = w / w % × 10 × d M 2 = 22 × 10 × 1.253 342 = 0.805 M E 2 = eq.mass of solute M 2 = molar ass of solute Molality = w / w % × 1000 M 2 100 - w / w % Molality = 22 × 1000 342 ( 100 - 22 ) = 0.825 m

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