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. Calculate the weight of the precipitate of `BaCO_(3)(s)` obtained from 50 " mL of " the above test solution. `(Ba=137,C=12,O=16,Mw(BaCO_(3))=137+12+3xx16=197gmol^(-1))`

Options

A. 3.94 g
B. 0.394 g
C. 1.97 g
D. 0.197 g

Explanation

To solve the problem step by step, we will follow the process of calculating the weight of the precipitate of BaCO₃ obtained from the KOH solution after CO₂ absorption.

### Step 1: Calculate the milliequivalents of KOH in the solution
The volume of KOH solution used is 500 mL and the concentration is 0.1 M.

\[ \text{Milliequivalents of KOH} = \text{Volume (mL)} \times \text{Molarity (mol/L)} \times \text{n factor} \]
For KOH, the n factor is 1.

\[ \text{Milliequivalents of KOH} = 500 \, \text{mL} \times 0.1 \, \text{mol/L} \times 1 = 50 \, \text{mEq} \]

### Step 2: Calculate the milliequivalents of HCl used for neutralization
The volume of HCl used is 30 mL and the concentration is 0.1 M.

\[ \text{Milliequivalents of HCl} = \text{Volume (mL)} \times \text{Molarity (mol/L)} \times \text{n factor} \]
For HCl, the n factor is also 1.

\[ \text{Milliequivalents of HCl} = 30 \, \text{mL} \times 0.1 \, \text{mol/L} \times 1 = 3 \, \text{mEq} \]

### Step 3: Calculate the milliequivalents of KOH that reacted with HCl
Since 3 mEq of HCl neutralized 3 mEq of KOH, we can find the remaining KOH.

\[ \text{Remaining KOH} = \text{Initial KOH} - \text{KOH reacted with HCl} \]
\[ \text{Remaining KOH} = 50 \, \text{mEq} - 3 \, \text{mEq} = 47 \, \text{mEq} \]

### Step 4: Determine the milliequivalents of CO₂ absorbed
The remaining KOH will be equal to the milliequivalents of CO₂ absorbed, as KOH reacts with CO₂ to form carbonate ions.

\[ \text{Milliequivalents of CO₂} = 47 \, \text{mEq} \]

### Step 5: Calculate the milliequivalents of BaCO₃ formed
The milliequivalents of CO₃²⁻ formed from CO₂ will also be equal to the milliequivalents of BaCO₃ formed.

\[ \text{Milliequivalents of BaCO₃} = 47 \, \text{mEq} \]

### Step 6: Calculate the weight of BaCO₃ precipitate
The equivalent weight of BaCO₃ is calculated using its molecular weight and n factor.

\[ \text{Molecular weight of BaCO₃} = 197 \, \text{g/mol} \]
\[ \text{Equivalent weight of BaCO₃} = \frac{\text{Molecular weight}}{\text{n factor}} = \frac{197}{2} = 98.5 \, \text{g/equiv} \]

Now, we can calculate the weight of BaCO₃ precipitate using the milliequivalents.

\[ \text{Weight of BaCO₃} = \text{Milliequivalents} \times \text{Equivalent weight} \times 10^{-3} \]
\[ \text{Weight of BaCO₃} = 47 \, \text{mEq} \times 98.5 \, \text{g/equiv} \times 10^{-3} = 4.63 \, \text{g} \]

### Final Answer
The weight of the precipitate of BaCO₃ obtained is **4.63 g**.

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