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# The equivalent volume of ${\text{C}}{{\text{O}}_{\text{2}}}$ in the following reaction is${\text{2NaHC}}{{\text{O}}_{\text{3}}} \to {\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O + C}}{{\text{O}}_{\text{2}}}$A) 22.4 litersB) 11.2 litersC) 5.6 litersD) 22 liters

Last updated date: 02nd Aug 2024
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Hint: At STP volume of one mole of any gas is 22.4 L. Equivalent volume is the ratio of standard volume to factor n. Factor n indicates the number of replaceable ${{\text{H}}^{\text{ + }}}$ions.

Formula Used: ${\text{Equivalent volume = }}\dfrac{{{\text{Standard volume}}}}{{{\text{n - factor}}}}$

The chemical reaction given to us is
${\text{2NaHC}}{{\text{O}}_{\text{3}}} \to {\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O + C}}{{\text{O}}_{\text{2}}}$
From the balanced chemical reaction, we can say that in this reaction 2 moles of sodium bicarbonate
(${\text{NaHC}}{{\text{O}}_{\text{3}}}$) dissociates into 1 mole of sodium carbonate (${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$), 1 mole of water (${{\text{H}}_{\text{2}}}{\text{O}}$) and 1 mole of carbon dioxide gas (${\text{C}}{{\text{O}}_{\text{2}}}$).
To calculate the equivalent volume of ${\text{C}}{{\text{O}}_{\text{2}}}$ gas we need to determine the standard volume of ${\text{C}}{{\text{O}}_{\text{2}}}$ gas and n-factor.
We know that At the STP volume of one mole of any gas is 22.4L. So the standard volume of ${\text{C}}{{\text{O}}_{\text{2}}}$ gas is 22.4L.
To determine the n-factor indicates the number of replaceable ${{\text{H}}^{\text{ + }}}$ions of acid. In this reaction, we can see that 2 moles of ${\text{NaHC}}{{\text{O}}_{\text{3}}}$contain 2 moles of replaceable ${{\text{H}}^{\text{ + }}}$ions. Hence the value of the n-factor is 2.
Now, using the standard volume 22.4L and n-factor 2 we can calculate the equivalent volume of ${\text{C}}{{\text{O}}_{\text{2}}}$ gas as follows:
${\text{Equivalent volume = }}\dfrac{{{\text{Standard volume}}}}{{{\text{n - factor}}}}$
${\text{Equivalent volume = }}\dfrac{{{\text{22}}{\text{.4L}}}}{{\text{2}}} = {\text{11}}{\text{.2L}}$
Thus, the equivalent volume of ${\text{C}}{{\text{O}}_{\text{2}}}$ gas for the given reaction is 11.2 L.

Hence, the correct answer is option (B) 11.2 liters.

Note: Acid is the species that donate ${{\text{H}}^{\text{ + }}}$ ions in an aqueous solution. Basicity of the acid indicates a number of replaceable ${{\text{H}}^{\text{ + }}}$ ions present in acid. While n-factor indicates an actual number of ${{\text{H}}^{\text{ + }}}$ ions replaced during the reaction. So in the case of diprotic, triprotic acids n-factor might be different from the basicity of acids.