MCQ
$KMn{O_4}$ reacts with ferrous ammonium sulphate according to the equation

$MnO_4^ - + 5F{e^{2 + }} + 8{H^ + } \to M{n^{2 + }} + 5F{e^{3 + }} + 4{H_2}O$,

here $10\, ml$ of $0.1\, M$ $KMn{O_4}$ is equivalent to

  • A
    $20\, ml$ of $ 0.1\, M$ $FeS{O_4}$
  • B
    $30\, ml$ of $0.1 \,M$ $FeS{O_4}$
  • C
    $40\, ml$ of $0.1\, M$ $FeS{O_4}$
  • $50\, ml$ of $0.1\, M$ $FeS{O_4}$

Answer

Correct option: D.
$50\, ml$ of $0.1\, M$ $FeS{O_4}$
d
(d) $KMn{O_4} = $ Mohr salt

$\frac{{{M_1}{V_1}}}{{{n_1}}} = \frac{{{M_2}{V_2}}}{{{n_2}}}$;

$\frac{{0.1 \times 10}}{1} = \frac{{{M_2}{V_2}}}{5}$;

${M_2}{V_2} = 5$

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