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(C) The reaction involves the formation of a white precipitate $Q$ when a coloured solution $P$ reacts with $BaCl_2$.$1$. $Na_2CrO_4$ is a yellow solu

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$CuSO_4$

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(C) The reaction involves the formation of a white precipitate $Q$ when a coloured solution $P$ reacts with $BaCl_2$.$1$. $Na_2CrO_4$ is a yellow solution,but it forms a yellow precipitate $BaCrO_4$,not white.$2$. $ZnSO_4$ is a colourless solution,so it does not fit the criteria of $P$ being a coloured solution.$3$. $CuSO_4$ is a blue (coloured) solution. The reaction is: $CuSO_{4(aq)} + BaCl_{2(aq)} 	o BaSO_{4(s)} \downarrow_{(white)} + CuCl_{2(aq)}$. Here,$BaSO_4$ is a white precipitate and $CuCl_2$ is a coloured (greenish-blue) solution. This matches the given conditions.$4$. $AgNO_3$ is a colourless solution.Therefore,the correct salt $P$ is $CuSO_4$.

Column-$I$	Column-$II$
$P$. $NaBr + \text{Conc. } H_2SO_4 \rightarrow$	$A$. Colourless gas
$Q$. $Na_2S + \text{dil. } HCl \rightarrow$	$B$. Brown colour
$R$. $NaNO_2 + \text{dil. } HCl \rightarrow$	$C$. Rotten egg smell
$S$. $NaNO_3 + \text{Conc. } H_2SO_4 \rightarrow$	$D$. Paramagnetic

$P_{(Coloured \ Solution)} + BaCl_2 \to Q \downarrow_{(White)} + R_{(Coloured \ Solution)}$
Then salt $P$ in above reaction is

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$P_{(Coloured \ Solution)} + BaCl_2 \to Q \downarrow_{(White)} + R_{(Coloured \ Solution)}$Then salt $P$ in above reaction is