Molarity of $N a_{2} C O_{3}$ solution formed in the above question will be:

0.0244

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0.0112

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0.1500

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0.1276

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Solution

The correct option is B $0.0112$ M
The reaction is as follows:
$2 N a O H (2 m o l) + C O_{2} (1 m o l) \to N a_{2} C O_{3} (1 m o l) + H_{2} O$
$1$ mol $C O_{2}$ reacts with $2$ mol $N a O H$ .
$∴ 11.2 \times 10^{- 3} m o l C O_{2}$ reacts with $0.0224$ moles $N a O H$ at start $= 0.15$ mol.
$N a O H$ unreacted $= 0.150 - 0.224 = 0.1276$ in $1$ L.
$[N a_{2} C O_{3}] = [C O_{2}] = 0.0112$ mol per L $= 0.0112$ M.

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