Question

How many grams of dibasic acid ( mol.wt 200 ) should be present in 100 ml of aqueous solution to give strength of 0.1 N?

Answer 1

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Dear ayesha

Normality =Gram equivalent of the soluteVolume of the solution in litreVolume of the solution in litre = 0.1 LGram equivalent of the solute =Mass of the solute in gramEquivalent mass of the soluteSolute = Dibasic acidEquivalent mass of acid =Molecular mass of acidBasicity=2002Basicity for dibasic acid = 2Normality =Mass of the solute in gramEquivalent mass of the solute×Volume of the solution in litreNormality = 0.10.1 =Mass of the solute acid in gram200/2 ×0.1Mass of the solute acid in gram =100 ×0.1×0.1 =1g AnswerRegards

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Answer 2

n-factor of a dibasic acid = 2

Molarity = Normality/n-factor = 0.1/2 = 0.05

Number of moles = Molarity x Volume(in litres) = 0.05 x 0.1 = 0.005

→ Number of moles = given weight / molecular weight

→ 0.005 = given weight / 200

→ given weight = 0.005 x 200 = 1 gram

Thus, 1 gram of Dibasic acid should be present in 100ml of aqueous solution to prepare a 0.1N solution.

Hope it helps!!