Use Na = 23, S = 32, O = 16. To a solution of Na2SO3, 25 mL of 0.1 M acidified KMnO4 was added (sulfite -> sulfate; MnO4- -> Mn^2+). The unreacted KMnO4 was just reduced by 10 mL of 0.5 M Fe^2+ (Fe^2+ -> Fe^3+). The mass of Na2SO3 in the solution was:

JEE Advanced levelSingle correct optionDifficulty 3 of 3rx146

Use Na = 23, S = 32, O = 16. To a solution of $Na X_{2} SO X_{3}$ , $25 mL$ of $0.1 M$ acidified $KMnO X_{4}$ was added (sulfite $\to$ sulfate; $MnO X_{4} X^{-} Mn X^{2 +}$ ). The unreacted $KMnO X_{4}$ was just reduced by $10 mL$ of $0.5 M$ $Fe X^{2 +}$ ( $Fe X^{2 +} Fe X^{3 +}$ ). The mass of $Na X_{2} SO X_{3}$ in the solution was:

Total

KMnO X_{4}

= 25 \times 0.1 \times 5 = 12.5

meq. Excess

= 10 \times 0.5 \times 1 = 5

meq. Used by sulfite

= 7.5

meq. Sulfite has

n = 2

, so

3.75

mmol

\times 126 = 0.4725 g