Question

to prepare 100ml of M/20 mohr's salt solution as standard solution, find out the molarity and strength of given kmno4 solution​

Answer 1

Final answer: (a) Prepared 100 mL of $\frac{M}{20}$ standard solution Mohr's salt solution.  

(b) (I) Molarity of $KMnO_{4}$ = $\frac{1}{X}$ M  

    (II)Strength of $KMnO_{4}$ =  $\frac{158.034}{X}$  $gL^{-1}$

Given that: We are given Mohr's salt.

To find: We have to find the answers of,

(a) Preparation of 100 mL of  $\frac{M}{20}$ Mohr's salt solution as standard solution.

(b) Molarity and strength of given $KMnO_{4}$ solution.​

Explanation:

a) Preparation of 100 mL of $\frac{M}{20}$ standard solution of Mohr’s salt solution.

• The quantity of Mohr’s salt required for the 100 mL of the solution having a molarity of $\frac{M}{20}$ can be calculated as follows.

Chemical formula of Mohr’s salt = $FeSO_{4}(NH_{4})_{2}.6H_{2}O$

The molar mass of Mohr’s salt = 392.14   $\frac{g}{mol}$

$Molarity of Mohr^{'}s salt = \frac{Number of moles of Mohr^{'}s salt}{Volume of solution in liters} \\\\ Molarity of Mohr^{'}s salt = \frac{(\frac{Weight of Mohr^{'}s salt}{Molar mass of Mohr^{'}s salt})}{Volume of solution in liters}$

$\frac{1}{20} = \frac{Weight of Mohr^{'}s salt}{392.14} * \frac{100}{1000}$

$Weight of Mohr^{'}s salt =$ $[ (Molarity of Mohr^{'}s salt)$ * $(Molar mass of Mohr^{'}s salt)$ * $(Volume of solution in liters)]$

$Weight of Mohr^{'}s salt = [\frac{1}{20} * (392.14) * \frac{100}{1000}] = 1.9607 g$

Mohr salt required for preparing 100 mL of $\frac{M}{20}$ Mohr’s salt solution

       = 1.9607 g

Procedure:

1. Weigh an empty glass bottle using a chemical balance.

2. Weigh accurately 1.9607 g of Mohr’s salt using a chemical balance.

3. Transfer the Mohr’s salt in to 100 mL standard flask with the help of funnel.

4. Wash the funnel with distilled water without removing it from the flask.

5. Make up the solution to 100 mL using distilled water. Shake well.

This solution is $\frac{M}{20}$ standard solution of Mohr’s salt.

(b) (I) Molarity of $KMnO_{4}$ solution:

• From the titration values X mL of given $KMnO_{4}$ solution is equivalent to 20ml of $\frac{M}{20}$ Mohr’s salt solution.

• We know that, $M_{1}V_{1} = M_{2}V_{2}$

Here,  $Molarity of Mohr^{'}s salt * Volume of Mohr’s salt$

             $= Molarity of KMnO_{4} * Volume of KMnO_{4}$  

$Molarity of KMnO_{4} =$  $\frac{Molarity of Mohr^{'}s salt * Volume of Mohr’s salt}{Volume of KMnO_{4}}$

$Molarity of KMnO_{4} =\frac{(\frac{1}{20} * 20)}{X}$

                                 $= \frac{1}{X}$     

• $Molarity of KMnO_{4} = \frac{1}{X} M$

(b) (II) Strength of $KMnO_{4}$ solution:

• We know that  $Strength = Molarity * Molar mass$

• Here,

$Strength of KMnO_{4} = Molarity of KMnO_{4} * Molar mass of KMnO_{4}$

• $Molarity of KMnO_{4} = \frac{1}{X}M$

• $Molar mass of KMnO_{4}=158.034 \frac{g}{mol}$  

• $Strength of KMnO_{4} =\frac{1}{X}* 158.034 =(\frac{158.034}{X}) gL^{-1}$

To know more about the concept please go through the links

https://brainly.in/question/48446869

https://brainly.in/question/50972720

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