Question:
Equivalent weights of \(KMnO_4\) and \(K_2Cr_2O_7\) in acidic medium are respectively. (Molecular weight of \(KMnO_4=M_A\) and Molecular weight of \(K_2Cr_2O_7=M_B\))
Equivalent weights of \(KMnO_4\) and \(K_2Cr_2O_7\) in acidic medium are respectively. (Molecular weight of \(KMnO_4=M_A\) and Molecular weight of \(K_2Cr_2O_7=M_B\))
Show Hint
In acidic medium:
\[
KMnO_4 \rightarrow n=5
\]
\[
K_2Cr_2O_7 \rightarrow n=6
\]
Always calculate equivalent weight using:
\[
\text{Equivalent Weight}=\frac{\text{Molecular Weight}}{n\text{-factor}}
\]
Updated On: Jun 26, 2026
- \(\dfrac{M_A}{3},\ \dfrac{M_B}{6}\)
- \(\dfrac{M_A}{6},\ \dfrac{M_B}{5}\)
- \(\dfrac{M_A}{3},\ \dfrac{M_B}{5}\)
- \(\dfrac{M_A}{5},\ \dfrac{M_B}{6}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Recall the formula for equivalent weight.
For an oxidizing or reducing agent, \[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{n\text{-factor}} \] where \(n\)-factor is the number of electrons gained or lost per molecule.
Step 2: Determine the \(n\)-factor of \(KMnO_4\) in acidic medium.
In acidic medium, \[ MnO_4^- \rightarrow Mn^{2+} \] The oxidation state of Mn changes from \[ +7 \rightarrow +2 \] Thus, electrons gained: \[ n=5 \] Therefore, \[ \text{Equivalent weight of } KMnO_4 = \frac{M_A}{5} \]
Step 3: Determine the \(n\)-factor of \(K_2Cr_2O_7\) in acidic medium.
In acidic medium, \[ Cr_2O_7^{2-}\rightarrow 2Cr^{3+} \] The oxidation state of Cr changes from \[ +6 \rightarrow +3 \] Each chromium gains \[ 3 \] electrons. Since there are two chromium atoms, \[ n=2\times3=6 \] Therefore, \[ \text{Equivalent weight of } K_2Cr_2O_7 = \frac{M_B}{6} \]
Step 4: Final conclusion.
Hence, \[ \boxed{ \text{Equivalent weights} = \frac{M_A}{5},\ \frac{M_B}{6} } \] Therefore, the correct option is \[ \boxed{(4)} \]
For an oxidizing or reducing agent, \[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{n\text{-factor}} \] where \(n\)-factor is the number of electrons gained or lost per molecule.
Step 2: Determine the \(n\)-factor of \(KMnO_4\) in acidic medium.
In acidic medium, \[ MnO_4^- \rightarrow Mn^{2+} \] The oxidation state of Mn changes from \[ +7 \rightarrow +2 \] Thus, electrons gained: \[ n=5 \] Therefore, \[ \text{Equivalent weight of } KMnO_4 = \frac{M_A}{5} \]
Step 3: Determine the \(n\)-factor of \(K_2Cr_2O_7\) in acidic medium.
In acidic medium, \[ Cr_2O_7^{2-}\rightarrow 2Cr^{3+} \] The oxidation state of Cr changes from \[ +6 \rightarrow +3 \] Each chromium gains \[ 3 \] electrons. Since there are two chromium atoms, \[ n=2\times3=6 \] Therefore, \[ \text{Equivalent weight of } K_2Cr_2O_7 = \frac{M_B}{6} \]
Step 4: Final conclusion.
Hence, \[ \boxed{ \text{Equivalent weights} = \frac{M_A}{5},\ \frac{M_B}{6} } \] Therefore, the correct option is \[ \boxed{(4)} \]
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