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ExamsJEE MainChemistry

6.023 x 10²² molecules are present in 10 g of a substance 'x'. The molarity of a solution made by dissolving 5 g of 'x' in 2 L of solution is _____ x 10⁻³ M. Find this value.

  1. 12.5
  2. 25
  3. 50
  4. 100

Correct answer: 25

Solution

Molar mass = 10 g / 0.1 mol = 100 g/mol; 5 g = 0.05 mol; molarity = 0.05/2 = 0.025 M = 25 x 10⁻³ M.

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