When $Y$ is treated with $KMnO_4$,then the final precipitate is of:
$'X' \xrightarrow[BaCl_{2} \text{ solution}]{} 'Y' \quad (\text{White ppt., insoluble in dil. } HCl)$
$'X' \xrightarrow[\text{Flame test}]{} \text{Golden yellow flame}$

  • A
    $BaSO_4$
  • B
    $MnSO_4$
  • C
    $K_2SO_4$
  • D
    \text{None of these}

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