1.A 10.0ml of vitamin C solution is titrated against standard 0.01300M KMnO 4 in

Question

1.A 10.0ml of vitamin C solution is titrated against standard 0.01300M KMnO4 in acidic solution.At the end point 12.73 ml of KMnO4 solution have been consumed.Calculate the molarity of vitamin C solution.

1.#mol of KMnO4

2.#mol of vitamin C

3.molarity of vitamin C solution.

2.For ammonia quantitation 20.00mL of 0.3120M HCI was added to 0.203g of a Amminenickel (II) compex compound .Resulting solution was titrated with 0.1105M NaOH.A volume of 10.56mL of NaOH solution was required to reach an end point in titration.Calculate mas% of NH3 (MM=17.00g/mol) in the complex compound .

Explanation / Answer

Answer :

Given :

Molarity of KMnO4 = 0.013 M and Volume required to reach end point = 12.73 mL.

Milimoles of KMnO4 = Molarity x Volume = 0.013 x 12.73 = 0.1655 milimoles.

Hence, Moles of KMnO4 = 0.1655 x 10-3 moles = 1.655 x 10-4 moles.

Moles of KMnO4 = 1.655 x 10-4 moles.

Now at End point

Moles of Vitamine C = Moles of KMnO4 = 1.655 x 10-4 moles.

Moles of Vitamine C = 1.655 x 10-4 moles.

Molarity of Vitamine C = say 'M'

Volume of Vitamine titrated = 10.0 mL = 10 x 10-3 L = 10-2 L

Then

Moles of Vitamin C = M x 10-2 moles.

But Moles of Vitamine C = 1.655 x 10-4 moles.

so,

M x 10-2 = 1.655 x 10-4.

M = 1.655 x 10-2 M

Molarity of Vitamine C = 1.655 x 10-2 M

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