Sodium thiosulfate (Na2S2O3, molar mass 158 g mol^-1; Na2S2O3.5H2O=248 g mol^-1) is oxidised by I2 to tetrathionate, and by excess Cl2 water to sulfate (S2O3^2- -> 2SO4^2-). Select the correct statement(s).

With I2, the n-factor of Na2S2O3 is 1

With Cl2, the n-factor of Na2S2O3 is 8

With Cl2, the equivalent mass of anhydrous Na2S2O3 is 19.75

With I2, the n-factor of Na2S2O3 is 1

With Cl2, the n-factor of Na2S2O3 is 8

With Cl2, the equivalent mass of anhydrous Na2S2O3 is 19.75

With I2, the equivalent mass of Na2S2O3.5H2O is 158

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3eq035

Sodium thiosulfate ( $Na X_{2} S X_{2} O X_{3}$ , molar mass $158 g mo l^{- 1}$ ; $Na X_{2} S X_{2} O X_{3} \cdot 5 H X_{2} O = 248 g mo l^{- 1}$ ) is oxidised by $I X_{2}$ to tetrathionate, and by excess $Cl X_{2}$ water to sulfate ( $S X_{2} O X_{3} X^{2 -} 2 SO X_{4} X^{2 -}$ ). Select the correct statement(s).

AWith $I X_{2}$ , the n-factor of $Na X_{2} S X_{2} O X_{3}$ is $1$
BWith $Cl X_{2}$ , the n-factor of $Na X_{2} S X_{2} O X_{3}$ is $8$
CWith $Cl X_{2}$ , the equivalent mass of anhydrous $Na X_{2} S X_{2} O X_{3}$ is $19.75$
DWith $I X_{2}$ , the equivalent mass of $Na X_{2} S X_{2} O X_{3} \cdot 5 H X_{2} O$ is $158$

With

I X_{2}

2 S X_{2} O X_{3} X^{2 -}

loses

2

^{-}

, so

n = 1

per ion. With

Cl X_{2}

: each S goes

+ 2 \to + 6

(4 e

^{-}

), two S give

n = 8

and equivalent mass

158/8 = 19.75

. The pentahydrate with

I X_{2}

has

n = 1

, so its equivalent mass is

248

, not

158