P_{(Coloured Solution)} + BaCl_2 to Q downar | vedclass

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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

Vedclass · Accepted Answer

$CuSO_4$

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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 $CuSO_4$.

$P_{(Coloured \ Solution)} + BaCl_2 \to Q \downarrow_{(White)} + R_{(Coloured \ Solution)}$
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