To a solution of a colourless efflorescent so | vedclass

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(C) The sodium salt is sodium thiosulphate $(Na_2S_2O_3 \cdot 5H_2O)$, which is a colourless efflorescent salt.
When dilute acid is added to it, it u

Vedclass · Accepted Answer

$Na_2S_2O_3$

Vedclass · Answer

(C) The sodium salt is sodium thiosulphate $(Na_2S_2O_3 \cdot 5H_2O)$, which is a colourless efflorescent salt.
When dilute acid is added to it, it undergoes disproportionation to form sulphur (white precipitate) and sulphur dioxide gas $(SO_2)$:
$Na_2S_2O_{3(s)} + 2H^+_{(aq)} ightarrow S_{(s)} + SO_{2(g)} + H_2O_{(\ell)} + 2Na^+_{(aq)}$
$SO_2$ gas turns acidified potassium dichromate solution green due to the reduction of $Cr(VI)$ to $Cr(III)$:
$K_2Cr_2O_{7(	ext{aq})} + 3SO_{2(	ext{g})} + H_2SO_{4(	ext{aq})} ightarrow K_2SO_{4(	ext{aq})} + Cr_2(SO_4)_{3(	ext{aq})} + H_2O_{(\ell)}$
Thus, the correct option is $C$.

List-$I$ (Anion)	List-$II$ (Gas evolved on reaction with dil. $H_2SO_4$)
$A. CO_3^{2-}$	$I$. Colourless gas which turns lead acetate paper black
$B. S^{2-}$	$II$. Colourless gas which turns acidified potassium dichromate solution green
$C. SO_3^{2-}$	$III$. Brown fumes which turn acidified $KI$ solution containing starch blue
$D. NO_2^{-}$	$IV$. Colourless gas evolved with brisk effervescence,which turns lime water milky

To a solution of a colourless efflorescent sodium salt, when dilute acid is added, a colourless gas is evolved along with the formation of a white precipitate. Acidified dichromate solution turns green when the colourless gas is passed through it. The sodium salt is

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