- A${K_2}Mn{O_4}$
- ✓$Mn{O_2}$
- C$Mn{(OH)_2}$
- D$M{n^{2 + }}$
$KMn{O_4}$ is first reduced to manganate and then to insoluble manganese dioxide. Colour changes first from purple to green and finally becomes colourless.
$2KMn{O_7} + 2KOH \to 2{K_2}Mn{O_4} + {H_2}O + O$
$2{K_2}Mn{O_4} + 2{H_2}O \to 2Mn{O_2} + 4KOH + 2O$
$\overline {2KMn{O_2} + {H_2}O\xrightarrow{{{\text{alkaline}}}}2Mn{O_2} + 2KOH + 3[O]} $
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$(i)$ $4.5\, mL$ $\quad (ii)$ $4.5\, mL$ $\quad (iii)$ $4.4\, mL$
$(iv)$ $4.4\, mL$ $\quad (v)$ $4.4\, mL$
If the volume of oxalic acid taken was $10.0 \,mL$ then the molarity of the $NaOH$ solution is .... $M.$ (Rounded-off to the nearest integer)


$[{R_H} = 1 \times {10^5}\,c{m^{ - 1}},\,h\, = 6.6\, \times {10^{ - 34}}\,Js\,\,c = 3\, \times \,{10^8}\,m{s^{ - 1}}]$