ALTERNATIVE TO PRACTICAL A A is a solution of KMnO\(_4\) containing 0.020 moldm\(^{-3}\). B is a solution of Fe\(^{2+}\) obtained by dissolving 3.8 g of iro...

(a) Volume of pipette used (Volume of B used) = 25.0 cm\(^3\)

Volume of A used =  \(\frac{Rough + 1st titre + 2nd titre}{3}\)

                             = \(\frac{18.20 + 18.10 + 18.20}{3}\)

                             = 18.166

                 V\(_A\) = 18.17 cm\(^3\)

(i) Concentration of B in moldm\(^{-3}\) 

  From the balanced equation :   MnO\(_4\)\(^-\)\(_{(aq)}\) +  5Fe\(^{2+}\)\(_{(aq)}\)  +  8H\(^+\)\(_{(aq)}\) → Mn\(^{2+}\)\(_{(aq)}\)  + Fe\(^{3+}\)\(_{(aq)}\) +  4H\(_2\)O\(_{(l)}\)   

C\(_A\) =   0.020 moldm\(^{-3}\)                                            N\(_A\) = 1

V\(_A\) =  18.17cm\(^3\)                                                        N\(_B\) = 5

C\(_B\) = ?

V\(_B\) =  25.0cm\(^3\)

Using   \(\frac{C_A V_A}{C_B V_B}\) = \(\frac{N_A}{N_B}\)

                     C\(_B\) = \(\frac{C_A V_A N_B}{N_A V_B}\)

                     C\(_B\) = \(\frac{ 0.02 X 18.17 X 5}{25 X 1}\)

                     C\(_B\) = 0.07268

                     C\(_B\) = 0.070 moldm\(^{-3}\)

(ii) Concentration of B in gdm\(^{-3}\)

     If  3.8 g of Fe granules was dissolved in 250 cm\(^3\) of B

          Xg of Fe granules will be dissolved in 1000 cm\(^3\) of B

          X = \(\frac{ 3.8 X 1000}{250}\) =  15.2 gdm\(^{-3}\)

    Concentration of B in gdm\(^{-3}\) will be 15.2 gdm\(^{-3}\)

(iii) Mass of Fe\(^{2+}\) in 250cm\(^3\) of B.

 To get the mass, we'll use the formula  n = \(\frac{mass}{Molar mass}\)

However, since do not know the number of moles, n , hence we'll need to obtain n, using  n =  \(\frac{CV}{1000}\) since we have C and V

So,  n =  \(\frac{{0.070}\times{250}}{1000}\) =  0.025 mole

Now, we can apply n = \(\frac{mass}{Molar mass}\) to obtain the mass, given that the Molar mass of Fe was given to be 56

                               n = \(\frac{mass}{Molar mass}\)

                               m = n x M = 0.018 x 56 = 1.02 g

Therefore, 1.02 g of Fe\(^{2+}\) in 250 cm\(^3\) of B.

(iv) Percentage of Fe\(^{2+}\) in the granules =  \(\frac{Mass of Fe^{2+}}{Total mass of Fe granules}\times 100\)

                                                                       = \(\frac{1.02}{3.8}\times 100\)

                                                                       = 26.89%

                                                                       = 26.9%