A mixture of CO and `CO_(2)` when treated with `I_(2)O_(5)` gives `I_(2)` vapours accoeding to the following equation:
`5CO+I_(2)O_(5)to5CO_(2)+I_(2)`
`I_(2)` vapour was separated and treated with `HClO_(4)` and the resultant `HIO_(4)` required 0.001 " mol of "hlycerol for complete oxidation. After treatment with `I_(2)O_(5)` and removal of `I_(2)` ,t he mixture was treated with excess of 0.1 N NaOH solution and finally this solution required 20 " mL of " 1 N HCl to reach end point using phenolphthalein as indicator, followed by methyl orange as indicator after the first end point, 10 " mL of " further HCl was consumed.
Q. The total volume of NaOH used in the problem is

Options

A. 30 Ml
B. 300 mL
C. 60 mL
D. 600 mL

Explanation

To solve the problem step by step, we will analyze the given information and apply stoichiometric principles.

### Step 1: Understand the Reaction
The reaction provided is:
\[ 5 \text{CO} + \text{I}_2\text{O}_5 \rightarrow 5 \text{CO}_2 + \text{I}_2 \]

From this reaction, we can see that 5 moles of CO react with 1 mole of I2O5 to produce 5 moles of CO2 and 1 mole of I2.

### Step 2: Define Variables
Let:
- \( n \) = number of millimoles of CO2 produced
- The initial number of millimoles of CO is \( n + 5 \) (since 5 moles of CO are produced from the reaction).

### Step 3: Determine NaOH Consumption
When the CO2 reacts with NaOH, it forms sodium carbonate (Na2CO3) and water. The stoichiometry shows that:
- 1 mole of CO2 requires 2 moles of NaOH.

Thus, the millimoles of NaOH required for \( n + 5 \) moles of CO2 is:
\[ 2(n + 5) \]

### Step 4: Calculate Remaining NaOH
Let \( B \) be the initial millimoles of NaOH. After the reaction with CO2, the remaining NaOH will be:
\[ B - 2(n + 5) \]

### Step 5: Analyze the Titration with HCl
The problem states that the solution required:
- 20 mL of 1 N HCl using phenolphthalein as an indicator.
- An additional 10 mL of HCl using methyl orange as an indicator.

The first endpoint (with phenolphthalein) corresponds to the neutralization of NaOH, while the second endpoint (with methyl orange) corresponds to the neutralization of Na2CO3.

### Step 6: Set Up Equations
From the first endpoint:
\[ B - 2(n + 5) + (n + 5) = 20 \]
This simplifies to:
\[ B - n - 5 = 20 \]
Thus:
\[ B - n = 25 \] (Equation 1)

From the second endpoint:
\[ n + 5 = 10 \]
Thus:
\[ n = 5 \] (Equation 2)

### Step 7: Solve for B
Substituting \( n = 5 \) into Equation 1:
\[ B - 5 = 25 \]
So:
\[ B = 30 \] millimoles of NaOH.

### Step 8: Calculate Volume of NaOH
Since the concentration of NaOH is 0.1 N, we can find the volume (V) using the formula:
\[ \text{Millimoles} = \text{Normality} \times \text{Volume (L)} \]
Thus:
\[ 30 = 0.1 \times V \]
\[ V = \frac{30}{0.1} = 300 \text{ mL} \]

### Final Answer
The total volume of NaOH used in the problem is **300 mL**.

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