A white precipitate insoluble in conc. $H N O_{3}$ is formed when the aqueous solution of $x$ in $N a O H$ is treated with barium chloride and bromine water.
So, $x$ is :

S O_{3}

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S O_{2}

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C O_{2}

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none of the above

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Solution

The correct option is B $S O_{2}$
When the aqueous solution of $S O_{2}$ in $N a O H$ is treated with barium chloride and bromine water, a white precipitate is formed which is insoluble in conc. $H N O_{3}$ .
$S O_{2} + 2 N a O H ⟶ N a_{2} S O_{3} + H_{2} O$
$N a_{2} S O_{3} + B a C l_{2} ⟶ B a S O_{3} + 2 N a C l$
$B r_{2} + H_{2} O ⟶ 2 H B r + [O]$
$B a S O_{3} + [O] ⟶ B a S O_{4}$
white precipitate
(insoluble in conc. $H N O_{3}$ )

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