43. 2 g sample of H202 solution containing y % H202 by weight require y mi of a KMnO4 solution for complete oxidation under acidic condition. The normality of the KMnO4 solution is : a. 10/34 b. 20/34 c. 10/17 d. 20/17

Step-by-step soch:

1. Mass percent: 2 g solution me y % $H_2O_2$ ka matlab $\text{mass of } H_2O_2 = \frac{y}{100}\times 2 = 0.02y\,\text{g}$

2. Equivalent weight of }$H\_2O\_2$: Acidic medium me $H_2O_2$ ( -1 \to 0 ) oxidation state me jaata hai, 2 e(^{-}) lost per molecule (\Rightarrow n\text{-factor}=2). Toh $\text{eq. wt.}=\frac{34}{2}=17$

3. Equivalents of }$H\_2O\_2$: $\text{equiv} = \frac{\text{mass}}{\text{eq.wt.}} = \frac{0.02y}{17}=\frac{y}{850}$

4. **Equivalents of }KMnO_4$**: Agar uska normality$N$hai aur volume$y$ mL hai, $\text{equiv}=N\times \frac{y}{1000}$

5. Equivalent concept: Reducing agent ke equivalents = Oxidising agent ke equivalents, isliye

$N\times \frac{y}{1000}=\frac{y}{850}$

6. Solve karo (y cancel ho jaata hai!)

$N=\frac{1000}{850}=\frac{20}{17}$

Bas! Option (d) sahi hai.

Imagine tumhare paas ek magic liquid hai jo bubbles (oxygen) bana sakta hai – yeh hai $H_2O_2$ (hydrogen peroxide). Ab tum ek purple potion (KMnO_4) laate ho jo is magic ko test karta hai. Question yeh pooch raha hai: agar 2 g is mixture ka kuch hissa (y percent) magic liquid hai, aur use test karne ke liye bilkul utna hi number (y mL) purple drops lagte hain, to purple potion kitna strong hai? Strong hone ka matlab hai uski normality – kitna reactive hai 1 liter mein.

Soch lo ki har drop = kuch soldiers. Dono liquids ke soldiers ek-dusre ko neutralize karte hain. Jab ginti barabar hoti hai, reaction khatam. Yeh ginti barabar hone ka funda hi equivalent concept hai. Agar tum numbers set kar do, tumhe pata chalega ki har liter purple potion me 20/17 soldiers hote hain.

Detailed Calculation

1. $\text{Mass of } H_2O_2 = \frac{y}{100}\times 2 = 0.02y\,\text{g}$

2. $\text{Eq. wt. of } H_2O_2 = 17$

3. $\text{Equivalents of } H_2O_2 = \frac{0.02y}{17}=\frac{y}{850}$

4. KMnO_4 volume = $y$ mL $=\frac{y}{1000}$ L, normality = $N$

   $\text{Equivalents of } KMnO_4 = N\times \frac{y}{1000}$

5. Titration condition:

   $N\times \frac{y}{1000}=\frac{y}{850}$

6. Simplify (y cancel):

   $N=\frac{1000}{850}=\frac{100}{85}=\frac{20}{17}$

Final Answer: $N = \frac{20}{17}$ (Option d)