Organic Compound

Chemistry Level 3

A blackish brown colored solid A A , which is an oxide of manganese, when fused with alkali metal hydroxide and an oxidizing agent such as K N O X 3 \ce{KNO3} , produces a dark green colored compound B B . Compound B B on disproportionation in neutral and acidic solution gives a purple colored compound C C .

What is the molecular mass of compound C C ?


The answer is 158.

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1 solution

( A ) M n O 4 2 M n O 2 ( A ) + 4 K O H + K N O 3 2 K 2 M n O 4 ( B ) + O 2 + 2 H 2 O \large \displaystyle (A) \rightarrow MnO_4\\ \large \displaystyle 2 \underbrace{MnO_2}_{(A)} \, + \, 4 KOH\, + \, KNO_3 \, \rightarrow \, 2 \underbrace{K_2MnO_4}_{(B)}\, + \, O_2 \, + \, 2 H_2O

On disproportionation in neutral and acidic solution gives a purple coloured compound,

The electrolytic oxidation is given by,

M n O 4 2 Manganate ion Alkali M n O 4 Permanganate ion ( C ) \large \displaystyle \underbrace{MnO_4^{2-}}_{\text{Manganate ion}} \xrightarrow{\text{Alkali}} \, \underbrace{MnO_4^{-}}_{\text{Permanganate ion }(C)}

( A ) = M n O 2 , ( B ) = K 2 M n O 4 , ( C ) = K M n O 4 . ( C ) Molecular mass = 39 + 55 + ( 16 × 4 ) = 39 + 55 + 64 = 158 . \large \displaystyle \therefore (A) = MnO_2, \, (B) = K_2MnO_4, \, (C) = KMnO_4.\\ \large \displaystyle \therefore (C) \text{ Molecular mass} = \color{#D61F06}{39} + \color{#20A900}{55} + \color{#69047E}{(16 \times 4)} = \color{#BA33D6}{39 + 55 + 64} = \color{#3D99F6}{\boxed{158}}.

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