Same quantity of electricity being used to liberate iodine (at anode) and a metal (at cathode). The mass of metal liberated at cathode is 0.617 g and the liberated iodine completely reduced by 46.3 ml of 0.124 M sodium thiosulphate solution. What is equivalent weight of metal?

23 g

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127 g

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108 g

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46 g

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Solution

The correct option is B 108 g
First calculate the mass of ${N a}_{2} S_{2} O_{3} . {5 H}_{2} O$
$\frac{0.124 m o l {N a}_{2} S_{2} O_{3} . {5 H}_{2} O}{1 L s o l u t i o n} \times 0.0463 L s o l u t i o n = 0.0057412 m o l {N a}_{2} S_{2} O_{3} . {5 H}_{2} O$

$= 0.0057412 m o l {N a}_{2} S_{2} O_{3} . {5 H}_{2} O \times 248.17 \frac{g {N a}_{2} S_{2} O_{3} . {5 H}_{2} O}{m o l {N a}_{2} S_{2} O_{3} . {5 H}_{2} O} = 1.42479 g {N a}_{2} S_{2} O_{3} . {5 H}_{2} O$

Now calculate the equivalent mass of metal.
$\frac{M a s s o f m e t a l}{E q u i v a l e n t w e i g h t o f m e t a l} = \frac{M a s s o f {N a}_{2} S_{2} O_{3} . {5 H}_{2} O}{E q u i v a l e n t w e i g h t o f {N a}_{2} S_{2} O_{3} . {5 H}_{2} O}$

$\frac{0.617 g}{E q u i v a l e n t w e i g h t o f m e t a l} = \frac{1.42479}{248.17}$

$E q u i v a l e n t w e i g h t o f m e t a l = 108 g .$

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