Question

EDTA Solation Calelate the Determine the normality of a solution of KMno, (FW-158.034) 17.935 g of primary standand KCr. (FW 294.202) was used to prepare 0.5000 L of this solution. b) Calculate the percent iron (At. Wt.-55.847 g/mole) in an unknown if a 0.9887 g sample reqaired 30.05 ml of the above standard KMnO, solation to titrate the iron as Fe in the sample. * Note: Answers should have the correct number of significant figures. +2 -1 +2 4. In the spectrophotometric determination of manganese in steel, a 0.250 gram sample of unknown steel is disolved and the manganese oxidized to Mno, This solution is then diluted to 500 ml in a volumetrie fask The quantity 0,60 gram of s The Spectronic 20, set at 540 nm to match t steel sample, certified to contain 0.25% Mn, is treated in exactly the same way. absorption band of the Mn04 ion, is adjusted so that the instrument reads 0% transmission with no sample in the Spectro ees ( trens mission with a distilled water blank in the instrument. The % transmission of the unknown solution en solution (ST(k)) is found to be 30,0%.

please answer #3 thank u

Answer 1

Following is the - complete Answer -&- Explanation - for the given - Question (i.e. Question - 3 ) - in Typed format...

$\Rightarrow$Answer:

1. Question - 3(a):   Normality of KMnO₄ solution: =  0.3252 N ( Answer )
2. Question - 3(b): Percent Iron in the unknown sample ( i.e. Iron compound ) =  93.2 % ( Answer )

$\Rightarrow$Explanation:

Following is the Explanation - for the above Answer(s) - in Typed format......

Question - 3 (a):

• Given:

1. Formula weight of KMnO₄ = 158.034 g/mole
2. Formula weight of K₂Cr₂O₇= 294.202 g/mole
3. Mass of K₂Cr₂O₇ sample = 7.9935 grams
4. Volume of the KMnO₄ - solution: V₁ = 0.500 L (i.e. Liters )

• ​​​​​​​Step - 1 :

​​​​​​​Moles of K₂Cr₂O₇in the solution = Mass / Molar mass = [ 7.9935 g ] / [ 294.202 g/mole ] = 0.0271 moles (approx.)

Since the above solution of K₂Cr₂O₇- has been used to prepare solution of KMnO₄, and each mole of K₂Cr₂O₇contains - two moles of potassium ( K ) , i.e. each mole of K₂Cr₂O₇will produce 2 moles of KMnO₄, thus -

Moles of KMnO₄ - in the solution will be : ( 0.0271 moles ) x 2   = 0.0542 moles - KMnO₄

Therefore - ( Since 0.5 L of solution contains 0.0542 moles of KMnO4 ... )

Molarity of the KMnO₄ - solution: M_KMnO4 = ( 0.0542 moles ) / (0.5 L ) = 0.1084 M (i.e. mole/L )

• Step - 2 :

​​​​​​​Since - in Neutral Medium : Number of Equivalents of KMnO₄ = 3  

$\therefore$Normality of the Solution ( N ) = Molarity ( M ) x number of equivalents ( i.e. for KMnO₄ solution )

= ( 0.1084 M ) x ( 3 equivalents ) = 0.3252 N

Therefore -

$\Rightarrow$Answer ( Question - 3 (a) ):

Normality of KMnO₄ solution: =  0.3252 N ( Answer )

Question - 3(b):

• Step - 1:

​​​​​​​As we have obtained earlier in the above solution, Molarity of the KMnO₄ solution :

      M_KMnO4 = 0.1084 M (i.e. mole/L )

• Step - 2:

​​​​​​​Therefore - Moles of KMnO₄ contained in 30.05 mL - of the above sample = ( 0.0305 L ) x ( 0.1084 mole/L )

= 0.0033 moles

• Step - 3:

​​​​​​​We know the following balanced chemical equation:

  $\Rightarrow$5 Fe ²⁺ + MnO₄^- + 8 H ⁺$\rightleftharpoons$ 5 Fe ³⁺ + Mn ²⁺ + 4 H₂O ------------------ ( Equation - 1 )

• Step - 4:

​​​​​​​Therefore:

According to Equation - 1 ---   0.0033 moles of KMnO₄ will neutralize: ( 0.0033 moles ) x 5 = 0.0165 moles of Fe ²⁺( i.e. the iron compound we are searching for ).

• Step - 5:

​​​​​​​As we have been informed in the problem, Mass of the iron ( Fe ²⁺ ) compound = 0.9887 g ( i.e. grams )

$\therefore$ We can say - Mass of 0.0165 moles - of the iron compound = 0.9887 g

$\therefore$Molar mass / FW = ( 0.9887 g / 0.0165 moles ) = 59.92 g/mole  $\approx$60.0 g / mole

• Step - 6:

​​​​​​​We know mass of Fe ²⁺ ion = 55.847 g/mole

$\therefore$   Mass of 0.0165 moles of Iron = ( 55.847 g/mole ) x ( 0.0165 mole ) = 0.9215 grams ( approx.)

Therefore:

• Step - 7:

​​​​​​​$\therefore$   Percent iron = [ ( 0.9215 g ) / ( 0.9887 g ) ] x 100 % = 93.2 % Iron

Therefore:

$\Rightarrow$Answer ( Question - 3 (b) ):

Percent Iron in the unknown sample ( i.e. Iron compound ) =  93.2 % ( Answer )