In the reaction: Na2S2O3 + 4Cl2 + 5H2O → Na2SO4 + H2SO4 + 8HCI, the equivalent weight of Na2S2O3 will be: (M = molecular weight of Na2S2O3)

Question

In the reaction: Na2S2O3 + 4Cl2 + 5H2O → Na2SO4 + H2SO4 + 8HCI, the equivalent weight of Na2S2O3 will be: (M = molecular weight of Na2S2O3) (A) M/4 (B) M/8 (C) M/1 (D) M/2

Tardigrade Admin · Accepted Answer

Na2+2S2O3 → Na2+6SO4 the total change in oxidation number = 4 × 2 = 8 ∴ ENa2S2O3 = (mol.wt./v f)=(M/8)

Q. In the reaction: $N a_{2} S_{2} O_{3} + 4 C l_{2} + 5 H_{2} O \to N a_{2} S O_{4} + H_{2} S O_{4} + 8 H C I,$ the equivalent weight of $N a_{2} S_{2} O_{3}$ will be: (M = molecular weight of $N a_{2} S_{2} O_{3}$ )

$M /4$

$M /8$

$M /1$

$M /2$

$N a_{2}^{+ 2} S_{2} O_{3} \to N a_{2}^{+ 6} S O_{4}$
the total change in oxidation number $= 4 \times 2 = 8$
$∴ E_{N a_{2} S_{2} O_{3}} = \frac{m o l . wt .}{v f} = \frac{M}{8}$

Q. In the reaction: Na2​S2​O3​+4Cl2​+5H2​O→Na2​SO4​+H2​SO4​+8HCI, the equivalent weight of Na2​S2​O3​ will be: (M = molecular weight of Na2​S2​O3​)

M/4

M/8

M/1

M/2