Step 1: Understanding the Concept:
This problem involves a redox reaction where hydrogen peroxide ($H_2O_2$) acts as an oxidizing agent. In an acidic medium, $H_2O_2$ oxidizes iodide ions ($I^-$) from potassium iodide (KI) to elemental iodine ($I_2$). By measuring the amount of iodine liberated, we can use stoichiometry to determine the original concentration of the hydrogen peroxide solution. Percentage strength usually refers to the mass of solute in grams per 100 mL of solution (% w/v).
Step 2: Key Formula or Approach:
1. Write and balance the chemical equation:
\[ H_2O_2 + 2KI + H_2SO_4 \rightarrow I_2 + K_2SO_4 + 2H_2O \]
2. Determine the molar relationship: 1 mole of $H_2O_2$ produces 1 mole of $I_2$.
3. Calculate moles of $I_2$ liberated.
4. Calculate mass of $H_2O_2$ reacted.
5. Calculate percentage strength (% w/v) = $\frac{\text{mass of } H_2O_2 (g)}{\text{Volume of solution (mL)}} \times 100$.
Step 3: Detailed Explanation:
From the balanced equation, we see that the mole ratio of $H_2O_2$ to $I_2$ is $1:1$.
First, let's calculate the moles of iodine liberated:
Molar mass of $I_2 = 2 \times 127 = 254$ g/mol.
\[ \text{Moles of } I_2 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{1.27 \text{ g}}{254 \text{ g/mol}} = 0.005 \text{ mol} \]
According to the stoichiometry, moles of $H_2O_2$ reacted must also be $0.005$ mol.
Now, convert moles of $H_2O_2$ to mass:
Molar mass of $H_2O_2 = (2 \times 1) + (2 \times 16) = 34$ g/mol.
\[ \text{Mass of } H_2O_2 = 0.005 \text{ mol} \times 34 \text{ g/mol} = 0.17 \text{ g} \]
This $0.17$ g of $H_2O_2$ is present in a $5.0$ $cm^3$ (which is $5.0$ mL) solution.
To find the percentage strength (% w/v):
\[ % \text{ strength} = \frac{0.17 \text{ g}}{5.0 \text{ mL}} \times 100 \]
\[ % \text{ strength} = 0.034 \times 100 = 3.4% \]
Step 4: Final Answer:
The percentage strength of the $H_2O_2$ solution is 3.4%. The correct option is (D).