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Q.2.75 g of Na2CO3 is present in 200 ml of Na2CO3 solution. Calculate the molarity of the solution.

Odisha ChseOdisha CHSE +2 Science Board Exam 2020Subjective· 2mImportance★★★★★
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Molarity = moles of solute / volume of solution in litres; compute moles of Na2CO3 from its mass and molar mass, then divide by the volume in litres.

Molar mass of Na2CO3 = 2(23) + 12 + 3(16) = 46 + 12 + 48 = 106 g/mol

Moles of Na2CO3 = given mass / molar mass = 2.75 g / 106 g/mol = 0.02594 mol

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