Question

\(x\) mL of \(0.05\,M\) \(KMnO_4\) solution is required to oxidise completely \(1.52\) g of \(FeSO_4\) in acidic medium. The value of \(x\) is \[ (\text{Atomic weights: } Fe=56,\ S=32,\ O=16) \]

• A. 40
• B. 20
• C. 30
• D. 50

Hint:

For \(KMnO_4\) in acidic medium: \[ n\text{-factor}=5 \] \[ N=M\times5. \] For redox titrations: \[ \boxed{N_1V_1=N_2V_2} \] or equivalently, \[ \boxed{\text{Equivalents of oxidant}=\text{Equivalents of reductant}} \]

Answer 1

The Correct Option is A

Concept: In acidic medium, \[ MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+}+4H_2O \] Thus, \[ 1\ \text{mol KMnO}_4 = 5\ \text{equivalents}. \] Also, \[ Fe^{2+} \rightarrow Fe^{3+}+e^-, \] so \[ 1\ \text{mol FeSO}_4 = 1\ \text{equivalent}. \] Hence, \[ \text{Equivalents of oxidant} = \text{Equivalents of reductant}. \]

Step 1: Calculate molar mass of \(FeSO_4\). \[ M(FeSO_4) = 56+32+4(16) \] \[ = 152\ \text{g mol}^{-1}. \]

Step 2: Calculate moles of \(FeSO_4\). \[ n = \frac{1.52}{152} = 0.01\ \text{mol}. \] Since \(FeSO_4\) has \(n\)-factor \(=1\), \[ \text{equivalents of }FeSO_4 = 0.01. \]

Step 3: Calculate normality of \(KMnO_4\). Given, \[ M=0.05. \] In acidic medium, \[ N=M\times5. \] \[ N=0.05\times5 =0.25. \]

Step 4: Use the equivalence relation. \[ N_1V_1=N_2V_2. \] \[ 0.25\times\frac{x}{1000} = 0.01. \] \[ x = \frac{0.01\times1000}{0.25}. \] \[ x=40. \]

Final Answer: \[ \boxed{x=40\ \text{mL}} \] \[ \boxed{\text{Answer = (A)}} \]