0-1 mol of KM nO 4 in acidic medium can oxidiseA- 0-5 mol of Mohr-s saltB- 0-25 mol of F eC 2 O 4C- 0-25 mol of H 2 C 2 O 4D- 0-025 mol of ferric alum

The correct options are A 0.5 mol of Mohr’s salt C 0.25 mol of H_2C_2O_4 MnO^ _4 + 8H^+ + 5e^ → Mn^2++4H_2 O E(KMnO_4) = molar mass/5(E=Eq. mass) Mohr's salt ...

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Byju's Answer

Standard XII

Chemistry

Depression in Freezing Point

0.1 mol of KM...

Question

0.1 mol of $K M n O_{4}$ in acidic medium can oxidise

0.5 mol of Mohr’s salt

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0.25 mol of $F e C_{2} O_{4}$

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0.25 mol of $H_{2} C_{2} O_{4}$

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0.025 mol of ferric alum

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Solution

The correct options are
A
0.5 mol of Mohr’s salt

C
0.25 mol of $H_{2} C_{2} O_{4}$

$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$

$E (K M n O_{4}) = \frac{molar mass}{5} (E=Eq. mass)$

Mohr's salt $(F e^{2 +}) \to F e^{3 +} + e^{-}$ ;

$E (F e C_{2} O_{4}) = \frac{molar mass}{3}$

$H_{2} C_{2} O_{4} \to 2 C O_{2} + 2 H^{+} + 2 e^{-}$

0.1 mol of $K M n O_{4} = 5 \times 0.1 = 0.5 eq$
0.5 mol of Mohr's salt $(F e^{2 +}) \to F e^{3 +} + e^{-} = 5 \times 0.1 = 0.5 eq$ ;
0.25 mol of $F e C_{2} O_{4} = 3 \times 0.25 = 0.75 eq$ ;
0.25 mol of $H_{2} C_{2} O_{4} = 2 \times 0.25 = 0.5 eq$ ;
Ferric alum $[(N H_{4})_{2} S O_{4} . F e_{2} (S O_{4})_{3} .24 H_{2} O]$

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0.1 mol of KMnO4 in acidic medium can oxidise

0.1 mol of $K M n O_{4}$ in acidic medium can oxidise