Q.

6.023 x 1022 molecules are present in 10 g of a substance ‘Y’. The molarity of a solution containing 5 g of substance ‘Y’ in 2 L solution is ______ x 10-3.

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answer is 25.

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Detailed Solution

6.023×1022 molecules in grams6.023×1023 molecules are present in 100 gmsMolecular mass of substance ‘Y’ = 100 gmsM=5100×2=0.025=25×10−3

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