25.0 mL of a Na2SO3 solution is treated with 50.0 mL of 0.0500 M iodine (SO3^2- + I2 + H2O -> SO4^2- + 2I- + 2H+). The unreacted iodine needs 18.0 mL of 0.100 M Na2S2O3. The concentration of Na2SO3 (126 g/mol) in g/L is:

JEE Advanced levelSingle correct optionDifficulty 3 of 3eq159

25.0 mL of a $Na X_{2} SO X_{3}$ solution is treated with 50.0 mL of 0.0500 M iodine ( $SO X_{3} X^{2 -} + I X_{2} + H X_{2} O SO X_{4} X^{2 -} + 2 I X^{-} + 2 H X^{+}$ ). The unreacted iodine needs 18.0 mL of 0.100 M $Na X_{2} S X_{2} O X_{3}$ . The concentration of $Na X_{2} SO X_{3}$ (126 g/mol) in g/L is:

Total

I X_{2}

= 0.0500

\times

50.0 = 2.50 mmol. Excess

I X_{2}

= (0.100

\times

18.0)/2 = 0.900 mmol, so

I X_{2}

used = 1.60 mmol =

SO X_{3} X^{2 -}

(1 : 1). Molarity = 1.60/25.0 = 0.064 M, concentration = 0.064

\times

126 = 8.06 g/L.