0 . 023 × 10 22 molecules are present in 10 g of a substance x ' ' . The molarity of a solution containing 5 g of substance 'x' in 2 L solution is _ _ _ _ _ _ _ _ _ × 10 − 3

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Number of mole of X = 6 . 022 × 10 22 6 . 022 × 10 23 = 10 Molar mass of x So molar mass of x = 100 g Molarity = 5 100 × 2 = 0 . 025 M Ans. 0 . 025 M M = 25 × 10 - 3

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