The weight of hypo ( Na 2 S 2 O 3 ⋅ 5 H 2 O ) required to make= 100 cm 3 of 0.2 N solution for use in the reaction 2 S 2 O 32-+ I 2 longrightarrow S 4 O 62-+2 I - will be

Question

The weight of hypo ( Na 2 S 2 O 3 ⋅ 5 H 2 O ) required to make= 100 cm 3 of 0.2 N solution for use in the reaction 2 S 2 O 32-+ I 2 longrightarrow S 4 O 62-+2 I - will be (A) 1.49 g (B) 2.98 g (C) 4.96 g (D) 8.94 g

Tardigrade Admin · Accepted Answer

According to the given reaction, 2 S 2 O 32- longrightarrow S 4 O 62-+2 e- ∴ Eq. wt. of Na 2 S 2 O 3 ⋅ 5 H 2 O =( text mol. wt /1)=(248/1)=248 1000 cm 3 of 1 N sol. require, Na 2 S 2 O 3 ⋅ 5 H 2 O =248 g ∴ 100 cm 3 of 0.2 N of the sol. require, Na 2 S 2 O 3 ⋅ 5 H 2 O =(248 × 0.2/1000) × 100=4.96 g

Q. The weight of hypo $(N a_{2} S_{2} O_{3} \cdot 5 H_{2} O)$ required to make= $100 c m^{3}$ of $0.2 N$ solution for use in the reaction $2 S_{2} O_{3}^{2 -} + I_{2} ⟶ S_{4} O_{6}^{2 -} + 2 I^{-}$ will be

$1.49 g$

$2.98 g$

$4.96 g$

$8.94 g$

Q. The weight of hypo (Na2​S2​O3​⋅5H2​O) required to make= 100cm3 of 0.2N solution for use in the reaction 2S2​O32−​+I2​⟶S4​O62−​+2I− will be

1.49g

2.98g

4.96g

8.94g

According to the given reaction, 2S2​O32−​⟶S4​O62−​+2e− ∴ Eq. wt. of Na2​S2​O3​⋅5H2​O=1 mol. wt ​=1248​=248 1000cm3 of 1N sol. require, Na2​S2​O3​⋅5H2​O=248g ∴100cm3 of 0.2N of the sol. require, Na2​S2​O3​⋅5H2​O =1000248×0.2​×100=4.96g

Q. The weight of hypo $(N a_{2} S_{2} O_{3} \cdot 5 H_{2} O)$ required to make= $100 c m^{3}$ of $0.2 N$ solution for use in the reaction $2 S_{2} O_{3}^{2 -} + I_{2} ⟶ S_{4} O_{6}^{2 -} + 2 I^{-}$ will be

$1.49 g$

$2.98 g$

$4.96 g$

$8.94 g$