In the reaction of sodium thiosulphate with I2 in aqueous medium the equivalent weight of sodium thiosulphate is equal to

Question

In the reaction of sodium thiosulphate with I2 in aqueous medium the equivalent weight of sodium thiosulphate is equal to (A) molar mass of sodium thiosulphate (B) the average of molar masses of Na2S2O3 and I2 (C) half the molar mass of sodium thiosulphate (D) molar mass of sodium thiosulphate × 2

Tardigrade Admin · Accepted Answer

Since 2S2O2-3 arrow S4 O2- 6 + 2e- is a 2 e- change, therefore, Eq. mass of Na2S2O3 = 2 × Mol. mass of Na2S2C3/2 = Mol. mass of Na2S2O3.

Q. In the reaction of sodium thiosulphate with $I_{2}$ in aqueous medium the equivalent weight of sodium thiosulphate is equal to

molar mass of sodium thiosulphate

the average of molar masses of $N a_{2} S_{2} O_{3}$ and $I_{2}$

half the molar mass of sodium thiosulphate

molar mass of sodium thiosulphate $\times$ 2

Since $2 S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -} + 2 e^{-}$ is a $2 e^{-}$ change, therefore, Eq. mass of $N a_{2} S_{2} O_{3} = 2 \times$ Mol. mass of $N a_{2} S_{2} C_{3} /2 =$ Mol. mass of $N a_{2} S_{2} O_{3}$ .

Q. In the reaction of sodium thiosulphate with I2​ in aqueous medium the equivalent weight of sodium thiosulphate is equal to

molar mass of sodium thiosulphate

the average of molar masses of Na2​S2​O3​ and I2​

half the molar mass of sodium thiosulphate

molar mass of sodium thiosulphate × 2

Since 2S2​O32−​→S4​O62−​+2e− is a 2e− change, therefore, Eq. mass of Na2​S2​O3​=2× Mol. mass of Na2​S2​C3​/2= Mol. mass of Na2​S2​O3​.

Q. In the reaction of sodium thiosulphate with $I_{2}$ in aqueous medium the equivalent weight of sodium thiosulphate is equal to

the average of molar masses of $N a_{2} S_{2} O_{3}$ and $I_{2}$

molar mass of sodium thiosulphate $\times$ 2

Since $2 S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -} + 2 e^{-}$ is a $2 e^{-}$ change, therefore, Eq. mass of $N a_{2} S_{2} O_{3} = 2 \times$ Mol. mass of $N a_{2} S_{2} C_{3} /2 =$ Mol. mass of $N a_{2} S_{2} O_{3}$ .