Q8 / 160 · AP EAMCET 2022 (06 Jul Shift 2) · Chemistry
Equivalent weights of $\mathrm{KMnO}_4$ and $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ in acidic medium are respectively.
(Molecular weight of $\mathrm{KMnO}_4=M_A$ and Molecular weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7=M_B$ ).
- A$\frac{M_A}{3} \cdot \frac{M_B}{6}$
- B$\frac{M_A}{6} \cdot \frac{M_B}{5}$
- C$\frac{M_A}{3} \cdot \frac{M_B}{5}$
- Correct answerD$\frac{M_A}{5} \cdot \frac{M_B}{6}$
Solution
Given that,
Molecular weight of $\mathrm{KMnO}_4=M_A$
$\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}$
$\therefore$ Equivalent weight of $\mathrm{KMnO}_4$
$=\frac{\text { Molecular weight }}{5}=\frac{M_A}{5}$
Similarly, molecular weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7=M_B$
$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+14 \mathrm{H}^{+}+6 e^{-} \longrightarrow 2 \mathrm{~K}^{+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}$
$\therefore$ Equivalent weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$
$=\frac{\text { Molecular weight }}{6}=\frac{M_B}{6}$
Hence, the correct answer is option: 'D'.