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Answer 1

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$\mathtt{\huge{\underline{\red{Answer\: :}}}}$

$\: \: \large \green{ \bf{ \underline{step \: by \: step \: explaination}}}$

$\: \implies \displaystyle \sf{the \: redox \: changes \: are : }$

$\: \bullet\bf{ {mn}^{ + 7} + 5e \longrightarrow \: {mn}^{2 + } }$

$\implies\displaystyle\sf{(0^-)_2 \longrightarrow 0^0_2+2e \bigg[E_(0_2)=\dfrac{M}{2}}\bigg]$

$\implies\displaystyle\sf{ \therefore Meq.\: of \: H_20_2=Meq.\: of \: Kmn0_4}$

$\implies\displaystyle\sf{\dfrac{w \times 1000}{\dfrac{34}{2}}=\dfrac{0.316}{\dfrac{158}{5}}\times 1000}$

$\: \implies \displaystyle \sf \therefore \: ekmn04 = \frac{158}{5}$

$\implies \displaystyle \sf \: \therefore \: wh202 = 0.17g$

$\: \\ \implies \displaystyle \sf \: \therefore \: percentage \: of \: purity \: of \: sample \: h202 = \frac{0.17 \times 100}{0.2} = 85 \: percent$