Question

JEE MAINS CHEMISTRY QUESTION SEPTEMBER 2020

Answer 1

ANSWER.

$\sf \to \: \% \: purity \: of \: h_{2} o_{2} \: = 85\%$

EXPLANATION.

$\sf \to \: 20 \: ml \: of \: 0.2 \: gram \: h_{2}o_{2} \: reacts \: completely \: with \: 0.316 \: gram \: kmno_{4} \\ \\ \sf \to \: kmno_{4} \: \: has \: \: oxidation \: \: \: number \: = 7 \\ \\ \sf \to \: mn {}^{ + 7} + 5 {e}^{ - } \to \: mn {}^{ + 2} \\ \\ \sf \to \: valency \: factor \: = v_{f} = 5 \\ \\ \sf \to \: h_{2}o_{2} \: has \: oxidation \: number \: = - 1 \\ \\ \sf \to \: o {}^{ - 1} _{2} \to \: o_{2} + 2e {}^{ - } \\ \\ \sf \to \: valency \: factor \: = v_{f} = 2$

$\sf \to \: M _{eq} h_{2}o_{2} = M_{eq}kmno_{4} \\ \\ \sf \to \: \frac{x}{ \frac{34}{2} } \times 1000 = \frac{0.316}{ \frac{158}{5} } \times 1000 \\ \\ \sf \to \: \frac{x}{17} \times 1000 = \frac{316}{ \frac{158 \times 1000}{5} } \times 1000 \\ \\ \sf \to \: \frac{x}{17} = \frac{1}{100}$

$\sf \to \: 100x = 17 \\ \\ \sf \to \: x = 0.17 \: gram \: \\ \\ \sf \to \: \% \: purity \: of \: h_{2}o_{2} \: = \frac{0.17}{0.2} \times 100 = 85\%$

Answer 2

$\mathfrak{\huge{\pink{ANSWER:-}}}$

$\begin{gathered}\sf :\implies \: 20 \: ml \: of \: 0.2 \: gram \: h_{2}o_{2} \: reacts \: completely \: with \: 0.316 \: gram \: kmno_{4} \\ \\ \sf :\implies \: kmno_{4} \: \: has \: \: oxidation \: \: \: number \: = 7 \\ \\ \sf \to \: mn {}^{ + 7} + 5 {e}^{ - } \to \: mn {}^{ + 2} \\ \\ \sf \implies \: valency \: factor \: = v_{f} = 5 \\ \\ \sf \to \: h_{2}o_{2} \: has \: oxidation \: number \: = - 1 \\ \\ \sf \to \: o {}^{ - 1} _{2} \to \: o_{2} + 2e {}^{ - } \\ \\ \sf \to \: valency \: factor \: = v_{f} = 2\end{gathered}$

$\begin{gathered}\sf :\implies \: M _{eq} h_{2}o_{2} = M_{eq}kmno_{4} \\ \\ \sf \to \: \frac{x}{ \frac{34}{2} } \times 1000 = \frac{0.316}{ \frac{158}{5} } \times 1000 \\ \\ \sf \to \: \frac{x}{17} \times 1000 = \frac{316}{ \frac{158 \times 1000}{5} } \times 1000 \\ \\ \sf \to \: \frac{x}{17} = \frac{1}{100}\end{gathered}$

$\begin{gathered}\sf :\implies \: 100x = 17 \\ \\ \sf :\implies \: x = 0.17 \: gram \: \\ \\ \sf :\implies \: \% \: purity \: of \: h_{2}o_{2} \: = \frac{0.17}{0.2} \times 100 = 85\%\end{gathered}$

Hence, $\orange{\underline{Purity \: = 85%}}$

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