When Y is treated with KMnO_4,then the final | vedclass

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(A) $1$. The flame test for '$X$' gives a golden yellow flame,which indicates the presence of sodium $(Na^+)$ ions.$2$. '$X$' reacts with $BaCl_2$ sol

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$BaSO_4$

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(A) $1$. The flame test for '$X$' gives a golden yellow flame,which indicates the presence of sodium $(Na^+)$ ions.$2$. '$X$' reacts with $BaCl_2$ solution to form a white precipitate '$Y$' that is insoluble in dilute $HCl$. This is a characteristic test for sulfate $(SO_4^{2-})$ ions,forming $BaSO_4$.$3$. Therefore,'$X$' is $Na_2SO_4$ and '$Y$' is $BaSO_4$.$4$. The question asks what happens when '$Y$' $(BaSO_4)$ is treated with $KMnO_4$. $BaSO_4$ is a stable,insoluble salt and does not react with $KMnO_4$ under normal conditions.$5$. Thus,there is no reaction,and the final precipitate remains $BaSO_4$.

When $Y$ is treated with $KMnO_4$,then the final precipitate is of:
$'X' \xrightarrow[BaCl_{2} \text{ solution}]{} 'Y' \quad (\text{White ppt., insoluble in dil. } HCl)$
$'X' \xrightarrow[\text{Flame test}]{} \text{Golden yellow flame}$

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When $Y$ is treated with $KMnO_4$,then the final precipitate is of:$'X' \xrightarrow[BaCl_{2} \text{ solution}]{} 'Y' \quad (\text{White ppt., insoluble in dil. } HCl)$$'X' \xrightarrow[\text{Flame test}]{} \text{Golden yellow flame}$