In a iodomeric estimation, the following reactions occur
`2Cu^(2+)+4i^(-) to Cu_(2)I_(2)+I_(2), I_(2)+2Na_(2)S_(2)O_(3) to 2NaI+Na_(2)S_(4)O_(6)`
0.12 mole of `CuSO_(4)` was adde to excess of KI solution and the liberated iodine required 120mL of hypo. The molarity of hypo soulution was:

Options

A. 2
B. `0.20`
C. `0.1`
D. `1.0`

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Understand the reactions
The reactions given are:
1. \(2Cu^{2+} + 4I^{-} \rightarrow Cu_2I_2 + I_2\)
2. \(I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6\)

From the first reaction, we see that 2 moles of \(Cu^{2+}\) react with 4 moles of iodide ions to produce 1 mole of iodine.

### Step 2: Calculate moles of \(Cu^{2+}\)
Given that 0.12 moles of \(CuSO_4\) were added, we can deduce that:
- The moles of \(Cu^{2+}\) present will also be 0.12 moles (since \(CuSO_4\) dissociates to give one \(Cu^{2+}\) ion).

### Step 3: Determine moles of iodine produced
From the stoichiometry of the first reaction:
- 2 moles of \(Cu^{2+}\) produce 1 mole of \(I_2\).
- Therefore, 0.12 moles of \(Cu^{2+}\) will produce:
\[ \text{Moles of } I_2 = \frac{0.12 \text{ moles of } Cu^{2+}}{2} = 0.06 \text{ moles of } I_2 \]

### Step 4: Calculate moles of \(Na_2S_2O_3\) used
From the second reaction:
- 1 mole of \(I_2\) reacts with 2 moles of \(Na_2S_2O_3\).
- Thus, 0.06 moles of \(I_2\) will react with:
\[ \text{Moles of } Na_2S_2O_3 = 0.06 \text{ moles of } I_2 \times 2 = 0.12 \text{ moles of } Na_2S_2O_3 \]

### Step 5: Calculate the volume of hypo solution used
We know that the volume of hypo (sodium thiosulfate solution) used is 120 mL, which is equivalent to 0.12 moles of \(Na_2S_2O_3\).

### Step 6: Calculate the molarity of the hypo solution
Molarity (M) is defined as moles of solute per liter of solution. We can calculate the molarity of the hypo solution as follows:
\[ \text{Molarity} = \frac{\text{Moles of } Na_2S_2O_3}{\text{Volume in liters}} = \frac{0.12 \text{ moles}}{0.120 \text{ L}} = 1 \text{ M} \]

### Final Answer
The molarity of the hypo solution is **1 M**.

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