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(i) A sample of MnSO_(4).4H_(2)Ois strongly heated in air . The residue is Mn_(3)O_(4) (ii) The residue is dissolved in 100 mL of 0.1N FeSO_(4)containing dilute H_(2)SO_(4) (iii) The solution reacts completely with 50 mL of KMnO_(4) solution . (iv) 25 mL of the KMnO_(4) solutionused in step(ii) requires 30 mL of 0.1 N FeSO_(4) solution for complete reaction . |
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Answer» Solution :m.e of 25 ML of `KMnO_(4)`solution = m.e of 30 mL of 0.1 N `FeSO_(4)` solution ` = 0.1 xx 30 = 3` ` :. ` m.e of 50 mL of `KMnO_(4)`solution = `2 xx 3 = 6` ` :. ` m.e of `FeSO_(4)`(remained which did not react with `Mn_(3)O_(4)`) = 6 Now , m.eof total `FeSO_(4)`solution = `0.1 xx 100 = 10 ` `:. ` m.eof `Mn_(3)O_(4) = 4` ` :. ` eq. of `Mn_(3)O_(4) = 4/1000` From the redoxreaction `{:(Mn_(3)O_(4)+Fe^(2+)to,3Mn^(2+)+Fe^(3+)),(+8,+6):}` Equivalent wt. of `Mn_(3)O_(4) = (" mol .wt")/(" change in ON PER mole") ` `= 229/2= 114.50` ` :." wt . of " Mn_(3)O_(4) = " equivalent " xx " eq.wt"` ` = 4/1000 xx 114.5 = 0.458 g ` As given in the problem`Mn_(3)O_(4)` is obtained by heating `MnSO_(4).4H_(2)O ` `MnSO_(4) . 4H_(2)O overset( DELTA) toMn_(3)O_(4)` or `(" wt . of " MnSO_(4).4H_(2)O)/(" mol.wt.of "MnSO_(4).4H_(2)O) = (3xx 0.458)/229` ` :. " wt . of " MnSO_(4) .4H_(2)O = (3xx 0.458 )/229 xx 223 ` ` = 1.338 g ` |
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