A substance on treatment with dilute H_2SO_4 | vedclass

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(C) The gas liberated is $SO_2$ (sulphur dioxide).$(i)$ $SO_2$ reacts with baryta water $(Ba(OH)_2)$ to form $BaSO_3$,which causes turbidity: $Ba(OH)_

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$SO_3^{2-}$

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(C) The gas liberated is $SO_2$ (sulphur dioxide).$(i)$ $SO_2$ reacts with baryta water $(Ba(OH)_2)$ to form $BaSO_3$,which causes turbidity: $Ba(OH)_2 + SO_2 	o BaSO_3 \downarrow + H_2O$.$(ii)$ $SO_2$ reduces acidified potassium dichromate $(K_2Cr_2O_7)$ to chromium$(III)$ sulphate,which is green in colour: $K_2Cr_2O_7 + 3SO_2 + H_2SO_4 	o K_2SO_4 + Cr_2(SO_4)_3 + H_2O$.These tests are characteristic of the sulphite ion $(SO_3^{2-})$.

$A$ substance on treatment with dilute $H_2SO_4$ liberates a colourless gas which produces:
$(i)$ turbidity with baryta water and
$(ii)$ turns acidified dichromate solution green.
The reaction indicates the presence of:

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$A$ substance on treatment with dilute $H_2SO_4$ liberates a colourless gas which produces:$(i)$ turbidity with baryta water and$(ii)$ turns acidified dichromate solution green.The reaction indicates the presence of: