In its reaction with iodine, 2S2O3^2- + I2 -> S4O6^2- + 2I-, the equivalent mass of Na2S2O3.5H2O (molar mass 248) is:

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In its reaction with iodine, $2 S X_{2} O X_{3} X^{2 -} + I X_{2} S X_{4} O X_{6} X^{2 -} + 2 I X^{-}$ , the equivalent mass of $Na X_{2} S X_{2} O X_{3} \cdot 5 H X_{2} O$ (molar mass 248) is:

A $124$
B $31$
C $158$
D $248$

Answer: D

Two thiosulphate ions lose 2 electrons in total, so each loses 1 electron and n-factor = 1. Equivalent mass = 248/1 = 248. 124 would need n = 2, and 31 would need n = 8; 158 is the mass of the anhydrous salt.

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In its reaction with iodine, 2SX2​OX3​X2−+IX2​​SX4​OX6​X2−+2IX−, the equivalent mass of NaX2​SX2​OX3​⋅5HX2​O (molar mass 248) is:

Answer: D

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In its reaction with iodine, $2 S X_{2} O X_{3} X^{2 -} + I X_{2} S X_{4} O X_{6} X^{2 -} + 2 I X^{-}$ , the equivalent mass of $Na X_{2} S X_{2} O X_{3} \cdot 5 H X_{2} O$ (molar mass 248) is: