In the reaction- I2-2Na2S2O3- 2NaI-Na2S4O6 - the equivalent weight of sodium thiosulphate is-

The correct option is C Molecular weight I_2+2Na_2S_2O_3→ 2NaI+Na_2S_4O_6 Change in oxidation state of sulphur = 0.5 . #160;As there are two s ...

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Oxidation and Reduction

In the reacti...

Question

In the reaction, $I_{2} + 2 N a_{2} S_{2} O_{3} \to 2 N a I + N a_{2} S_{4} O_{6}$ , the equivalent weight of sodium thiosulphate is:

M o l e c u l a r w e i g h t

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\frac{M o l e c u l a r w e i g h t}{2}

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\frac{M o l e c u l a r w e i g h t}{3}

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\frac{M o l e c u l a r w e i g h t}{4}

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H_{3} P O_{2} = \frac{m o l e c u l a r w e i g h t}{1}

H_{3} P O_{3} = \frac{m o l e c u l a r w e i g h t}{2}

H_{3} P O_{4} = \frac{m o l e c u l a r w e i g h t}{3}

(N a_{2} S_{2} O_{3} \cdot 5 H_{2} O)

2 N a_{2} S_{2} O_{3} + I_{2} \to 2 N a I + N a_{2} S_{4} O_{6}

2 {N H}_{3} ⟶ N_{2} + 2 {N H}_{3}

{N H}_{3}

= \frac{M o l e c u l a r w e i g h t}{y}

y

N H_{3}

N_{2} \to N H_{3}

N_{2}

= \frac{Molecular weight}{Number of electrons lost or gained per molecule}

I_{2} + 2 N a_{2} S_{2} O_{3} \to 2 N a I + N a_{2} S_{4} O_{6}

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