Air sample from an industrial area of Delhi, which is heavily polluted by `CO_(2)`, was collected and analysed. One such sample of 224 L of air measured at STP was passed through a 500 " mL of " 0.1 M KOH solution, where `CO_(2)(g)` was absorbed completely. 50 " mL of " the above solution was then treated with excess of `BaCl_(2)` solution where all the carbonate was precipitated as `BaCO_(4)(s)`. The solution was filtered off and the filtrate required 30 " mL of " 0.1 M HCl solution for neutralisation.
Q. ppm strength of `CO_(2)(g)` volume by volume `(" mL of " CO_(2)` per `10^(6)` L of air) is

Options

A. 224
B. 2240
C. 2350
D. 1000

Explanation

To solve the problem step by step, we will follow the procedure outlined in the video transcript.

### Step-by-Step Solution:

1. **Calculate the Initial Millimoles of KOH:**
\[ \text{Volume of KOH solution} = 500 \, \text{mL} = 0.5 \, \text{L} \]
\[ \text{Molarity of KOH} = 0.1 \, \text{M} \]
\[ \text{Millimoles of KOH} = \text{Volume (L)} \times \text{Molarity (mol/L)} \times 1000 \, \text{(to convert to millimoles)} \]
\[ = 0.5 \, \text{L} \times 0.1 \, \text{mol/L} \times 1000 = 50 \, \text{mmol} \]

**Hint:** Remember that 1 L = 1000 mL, and to convert moles to millimoles, multiply by 1000.

2. **Calculate the Unreacted KOH:**
\[ \text{Volume of HCl used} = 30 \, \text{mL} = 0.030 \, \text{L} \]
\[ \text{Molarity of HCl} = 0.1 \, \text{M} \]
\[ \text{Millimoles of HCl} = \text{Volume (L)} \times \text{Molarity (mol/L)} \times 1000 \]
\[ = 0.030 \, \text{L} \times 0.1 \, \text{mol/L} \times 1000 = 3 \, \text{mmol} \]
\[ \text{Unreacted KOH} = \text{Initial KOH} - \text{HCl used} = 50 \, \text{mmol} - 3 \, \text{mmol} = 47 \, \text{mmol} \]

**Hint:** The unreacted amount is simply the initial amount minus what was used in the reaction.

3. **Determine the Millimoles of CO2 Absorbed:**
The reaction between CO2 and KOH is given by:
\[ \text{CO2} + 2 \text{KOH} \rightarrow \text{K2CO3} + \text{H2O} \]
From the stoichiometry, 1 mmol of CO2 reacts with 2 mmol of KOH. Therefore:
\[ \text{Millimoles of CO2} = \frac{\text{Unreacted KOH}}{2} = \frac{47 \, \text{mmol}}{2} = 23.5 \, \text{mmol} \]

**Hint:** Pay attention to the stoichiometric coefficients in the balanced equation.

4. **Calculate the Volume of CO2 at STP:**
At STP, 1 mole of gas occupies 22.4 L. Thus:
\[ \text{Volume of CO2} = \text{Millimoles of CO2} \times \frac{22.4 \, \text{L}}{1000 \, \text{mmol}} = 23.5 \, \text{mmol} \times \frac{22.4 \, \text{L}}{1000} = 0.5264 \, \text{L} \]

**Hint:** Remember to convert millimoles to moles by dividing by 1000.

5. **Calculate the ppm of CO2:**
To find the ppm (parts per million) of CO2 in the air sample:
\[ \text{ppm} = \frac{\text{Volume of CO2 (L)}}{\text{Volume of air sample (L)}} \times 10^6 \]
\[ = \frac{0.5264 \, \text{L}}{224 \, \text{L}} \times 10^6 \]
\[ = 2350 \, \text{ppm} \]

**Hint:** The formula for ppm is a ratio of the solute volume to the solution volume, multiplied by \(10^6\).

### Final Answer:
The ppm strength of \(CO_2(g)\) in the air sample is **2350 ppm**.