`H_(2)O_(2)+2KIoverset(40% "yield")rarrI_(2)+2KOH`
`H_(2)O_(2)+2KMnO_(4)+3H_(2)SO_(4)overset(50% "yield")rarr K_(2)SO_(4)+2MnSO_(4)+3O_(2)+4H_(2)O`
150mL of `H_(2)O_(2)` sample was divided into two parts. First part was treated with KI and formed KOH required 200 mL of `M//2H_(2)SO_(4)` for neutralisation. Other part was treated with `KMnO_(4)` yielding 6.74 litre of `O_(2)` at 1 atm. and 273 K. Using % yield indicated find volume strength of `H_(2)O_(2)` sample used.

Options

A. 5.04
B. 10.08
C. 3.36
D. 33.6

Answer

To solve the problem step by step, we will analyze the reactions and calculate the required values systematically.

### Step 1: Calculate the moles of \( H_2SO_4 \) used for neutralization
Given that 200 mL of \( \frac{M}{2} H_2SO_4 \) is used for neutralization, we first convert the volume to liters:
\[ \text{Volume in liters} = \frac{200 \text{ mL}}{1000} = 0.2 \text{ L} \]
Now, calculate the molarity of \( H_2SO_4 \):
\[ \text{Molarity} = \frac{M}{2} = 0.5 \text{ M} \]
Now, we can find the moles of \( H_2SO_4 \):
\[ \text{Moles of } H_2SO_4 = \text{Molarity} \times \text{Volume} = 0.5 \times 0.2 = 0.1 \text{ moles} \]

### Step 2: Calculate the moles of \( KOH \) produced
From the reaction:
\[ H_2O_2 + 2KI \rightarrow I_2 + 2KOH \]
1 mole of \( H_2O_2 \) produces 2 moles of \( KOH \). Therefore, the moles of \( KOH \) produced can be calculated as:
\[ \text{Moles of } KOH = 2 \times \text{Moles of } H_2SO_4 = 2 \times 0.1 = 0.2 \text{ moles} \]

### Step 3: Calculate the moles of \( H_2O_2 \) used in the first reaction
Given the yield is 40%, we can find the moles of \( H_2O_2 \) used:
\[ \text{Moles of } H_2O_2 = \frac{\text{Moles of } KOH}{2} \times \frac{1}{0.4} = \frac{0.2}{2} \times \frac{1}{0.4} = 0.25 \text{ moles} \]

### Step 4: Calculate the moles of \( O_2 \) produced in the second reaction
Given that 6.74 L of \( O_2 \) is produced at STP (1 atm and 273 K), we can convert this volume to moles:
\[ \text{Moles of } O_2 = \frac{6.74 \text{ L}}{22.4 \text{ L/mol}} = 0.3 \text{ moles} \]

### Step 5: Calculate the moles of \( H_2O_2 \) used in the second reaction
From the reaction:
\[ H_2O_2 + 2KMnO_4 + 3H_2SO_4 \rightarrow K_2SO_4 + 2MnSO_4 + 3O_2 + 4H_2O \]
1 mole of \( H_2O_2 \) produces 3 moles of \( O_2 \). Thus, the moles of \( H_2O_2 \) used can be calculated as:
\[ \text{Moles of } H_2O_2 = \frac{\text{Moles of } O_2}{3} = \frac{0.3}{3} = 0.1 \text{ moles} \]

### Step 6: Calculate the total moles of \( H_2O_2 \) used
Now, we can find the total moles of \( H_2O_2 \) used:
\[ \text{Total moles of } H_2O_2 = 0.25 + 0.1 = 0.35 \text{ moles} \]

### Step 7: Calculate the molarity of \( H_2O_2 \) in the original 150 mL sample
Convert 150 mL to liters:
\[ \text{Volume in liters} = \frac{150 \text{ mL}}{1000} = 0.15 \text{ L} \]
Now, calculate the molarity:
\[ \text{Molarity of } H_2O_2 = \frac{\text{Total moles}}{\text{Volume}} = \frac{0.35}{0.15} \approx 2.33 \text{ M} \]

### Step 8: Calculate the volume strength of \( H_2O_2 \)
The volume strength of \( H_2O_2 \) is given by the formula:
\[ \text{Volume Strength} = 11.2 \times \text{Molarity} \]
Thus,
\[ \text{Volume Strength} = 11.2 \times 2.33 \approx 26.1 \text{ volume} \]

### Final Answer:
The volume strength of the \( H_2O_2 \) sample used is approximately **26.1 volume**.

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