The CO in a 20L sample of gas was converted to `CO_2` by passing the gas over iodine pentoxide heated to `150^@C`:
`I_2O_5(s)+5CO(g)to5CO_2(g)+I_2(g)`
The iodine distilled at this temperature and was collected in an absorber containing 10 mL of 0.011 M `Na_2S_2O_3`.The excess hypo was back titrated with 5 mL of 0.001 M `I_2` solution.
What must be the milligrams of Co in 1 L of the original gas sample ?

Options

A. 0.35 mg
B. 0.70 mg
C. 0.07 mg
D. 3.5 mg

Explanation

To solve the problem step by step, we will follow the outlined procedure based on the information provided in the question.

### Step 1: Calculate the millimoles of Na2S2O3 used
The volume of Na2S2O3 solution is given as 10 mL with a concentration of 0.011 M.

\[ \text{Millimoles of Na2S2O3} = \text{Volume (L)} \times \text{Molarity (mol/L)} = 0.010 \, \text{L} \times 0.011 \, \text{mol/L} = 0.00011 \, \text{mol} = 0.11 \, \text{mmol} \]

### Step 2: Calculate the millimoles of I2 used in back titration
The volume of I2 solution used for back titration is 5 mL with a concentration of 0.001 M.

\[ \text{Millimoles of I2} = \text{Volume (L)} \times \text{Molarity (mol/L)} = 0.005 \, \text{L} \times 0.001 \, \text{mol/L} = 0.000005 \, \text{mol} = 0.005 \, \text{mmol} \]

### Step 3: Calculate the millimoles of I2 liberated
To find the millimoles of I2 liberated, we subtract the millimoles of I2 used in back titration from the millimoles of Na2S2O3.

\[ \text{Millimoles of I2 liberated} = \text{Millimoles of Na2S2O3} - \text{Millimoles of I2 used} = 0.11 \, \text{mmol} - 0.005 \, \text{mmol} = 0.105 \, \text{mmol} \]

### Step 4: Relate I2 to CO using the stoichiometry of the reaction
From the reaction, we see that 1 mole of I2 corresponds to 5 moles of CO. Therefore, the millimoles of CO can be calculated as follows:

\[ \text{Millimoles of CO} = \left(\frac{5 \, \text{mmol CO}}{1 \, \text{mmol I2}}\right) \times 0.105 \, \text{mmol I2} = 0.525 \, \text{mmol CO} \]

### Step 5: Calculate the mass of CO in the original gas sample
The molar mass of CO is approximately 28 g/mol. To find the mass of CO in milligrams:

\[ \text{Mass of CO} = \text{Millimoles of CO} \times \text{Molar mass of CO} = 0.525 \, \text{mmol} \times 28 \, \text{g/mol} = 0.0147 \, \text{g} = 14.7 \, \text{mg} \]

### Step 6: Calculate the mass of CO in 1 L of the original gas sample
Since the original gas sample was 20 L, we need to find the concentration per liter:

\[ \text{Mass of CO in 1 L} = \frac{14.7 \, \text{mg}}{20} = 0.735 \, \text{mg} \]

### Final Answer
The mass of CO in 1 L of the original gas sample is **0.735 mg**.