Question

Match the following substances with their respective equivalent weights: 

• A. A - III; B - I; C - IV; D - II
• B. A - III; B - IV; C - I; D - II
• C. A - II; B - III; C - IV; D - I
• D. A - IV; B - II; C - III; D - I
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Hint:

To determine equivalent weight, always identify the oxidation state changes or replaceable H in acids/bases.

Answer 1

The Correct Option is A

Step 1: Understanding Equivalent Weight Formula Equivalent weight \( E \) is given by: \[ E = \frac{\text{Molar mass} (M)}{\text{n-factor}} \]
where \( n \)-factor is the number of replaceable hydrogen or electrons involved in the reaction. Step 2: Calculating Equivalent Weights
- \( Na_2CO_3 \) (Sodium Carbonate):
- Acts as a dibasic salt (providing 2 \( Na^+ \)), so \( n = 2 \).
- Equivalent weight = \( \frac{M}{2} \).
- (A - III). - \( KMnO_4 | H^+ \) (Acidic Medium):
- In acidic medium, Mn changes from +7 to +2, i.e., 5-electron transfer.
- Equivalent weight = \( \frac{M}{5} \).
- (B - I). - \( K_2Cr_2O_7 | H^+ \) (Acidic Medium):
- In acidic medium, Cr changes from +6 to +3, meaning a total 6-electron transfer.
- Equivalent weight = \( \frac{M}{6} \). - (C - IV).
- \( KMnO_4 | H_2O \) (Neutral/Basic Medium):
- In neutral or basic medium, Mn changes from +7 to +4, i.e., 3-electron transfer.
- Equivalent weight = \( \frac{M}{3} \).
- (D - II). Thus, the correct matching is A - III; B - I; C - IV; D - II. \bigskip