The eq. wt. of ${\rm{N}}{{\rm{a}}_2}{{\rm{S}}_2}{{\rm{O}}_3}$ as reductant, in the reaction, ${\rm{N}}{{\rm{a}}_2}{{\rm{S}}_2}{{\rm{O}}_3} + 5{{\rm{H}}_2}{\rm{O}} + 4{\rm{C}}{{\rm{l}}_2} \to 2{\rm{NaHS}}{{\rm{O}}_4} + 8{\rm{HCl}}:$

$\left( {{\rm{Mol}}{\rm{.wt}}{\rm{.}}} \right)/1$

NEET UG CHEMISTRY MCQ: REDOX REACTIONS

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Application +4 | -1

The eq. wt. of

\mathrm{N}{\mathrm{a}_2}{\mathrm{S}_2}{\mathrm{O}_3}

as reductant, in the reaction,

\mathrm{N}{\mathrm{a}_2}{\mathrm{S}_2}{\mathrm{O}_3} + 5{\mathrm{H}_2}\mathrm{O} + 4\mathrm{C}{\mathrm{l}_2} \to 2\mathrm{NaHS}{\mathrm{O}_4} + 8\mathrm{HCl}:

View answer and explanation

Correct option: D

(Mol. wt.)/8

Option D ((Mol. wt.)/8) is correct because the change in oxidation state of S in Na2S2O3 is from +2 to +6/2 = +3. Therefore, the n-factor is 2 * (3-2) * 2 = 2 * 1 * 2 = 8. Eq. wt = Mol. wt / n-factor = (Mol. wt.)/8

NEET UG CHEMISTRY MCQ: REDOX REACTIONS

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