A) sodium sulphite
B) sulphuric acid
C) sodium sulphate
D) sodium thiosulphite
Correct Answer: D
Solution :
Key Idea: The colour of reagent disappears when the products formed during reaction are colourless. \[\underset{\begin{smallmatrix} sodium \\ thiosulphite \end{smallmatrix}}{\mathop{2N{{a}_{2}}{{S}_{2}}{{O}_{3}}}}\,\,\,+\underset{\begin{smallmatrix} iodine \\ (purple\text{ }coloured) \end{smallmatrix}}{\mathop{{{I}_{2}}}}\,\,\,\xrightarrow{{}}\] \[\underset{\begin{smallmatrix} sodium\text{ }tetrathionate \\ (colourless) \end{smallmatrix}}{\mathop{N{{a}_{2}}{{S}_{4}}{{O}_{6}}}}\,\,\,\,\,\,+2Nal\]
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