Question

M/30 Mohr salt and find the molarity and strength 9f given kmno4 solution

Answer 1

Answer:

follows.

Molar mass of mohr’s salt = 392 g/mol

Mohr’s salt = x 0.05

Mohr’s salt = x 0.05 = 4.9 gm

Weigh an empty watch glass using a chemical balance.

Weigh accurately 4.9gm of Mohr’s salt in a chemical balance.

With the help of a funnel transfer the Mohr’s salt into the measuring flask.

Now wash the funnel with distilled water without removing the funnel from the flask.

Make the solution up to the marked point with distilled water and make sure the Mohr’s salt is fully dissolved.

This solution is 0.05N standard solution of Mohr’s salt.

(b) Titration of potassium permanganate solution against standard ferrous ammonium sulfate (Mohr’s salt) solution:

Wash and rinse the burette and pipette with distilled water and then rinse with the corresponding solution to be filled in them.

Rinse the burette with the potassium permanganate solution and fill the burette with potassium permanganate solution.

Fix the burette in the burette stand and place the white tile below the burette in order to find the endpoint correctly.

Rinse the pipette and conical flask with standard ferrous sulfate solution.

Pipette out 10ml of 0.05N standard Mohr’s salt solution into the conical flask.

Add a test tube full of sulfuric acid in order to prevent oxidation of manganese to form manganese dioxide.

Note down the initial reading in the burette before starting the titration.

Now start the titration, titrate against potassium permanganate solution and simultaneously swirl the solution in the flask gently.

Initially, the purple colour of KMnO4 is discharged with ferrous ammonium sulfate. The appearance of a permanent pink colour reveals the endpoint.

Repeat the titration until concordant values are obtained.

Note down the upper meniscus on the burette readings.

Record the reading in the observation table given below in order to calculate the molarity of KMnO4 given

Answer 2

Given, Mohr's salt i.e. $[(NH_{4})_{2}.Fe(SO_{4})_{2}.(H_{2}O)_{6}.$

Molarity $=\frac{1}{30}$

Molar mass $= 392 gm/mol$

We know, that Strength = Normality x Equivalent mass

and Normality = Molarity x n

As n for Mohr's salt is 1

$\therefore$ Normality = Molarity$=\frac{1}{30}N$

As n =1 , Equivalent mass = 392 gm/mol

$\therefore$ Strength = $\frac{1}{30} \times 392=13.067 gL^{-1}$

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