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Byju's Answer

Standard XII

Chemistry

Nitric Acid

Metal nitrate...

Question

Metal nitrate $(A)$ on heating decomposes, leaving a residue $(B)$ which goes into the solution with dil. $H C l$ . The solution of $(B)$ gives white ppt. with ammonium carbonate solution. Ppt. $C$ dissolves in dil. $H C l$ and solution is treated with potassium chromate to yellow ppt. $(D)$ . Solution $V$ with dil. $H_{2} S O_{4}$ also gives white ppt. $(E)$ insoluble in dil. $H C l$ and nitric acid. Precipitate $(E)$ is part of white pigment lithopone.

What is $(E)$ ?

B a S O_{4}

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M g S O_{4}

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C a S O_{4}

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N a_{2} S O_{4}

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Solution

The correct option is A $B a S O_{4}$
The sequence of reactions is given below.
$B a (N O_{3})_{2} h e a t - - \to B a O d i l H C l - --- \to B a C l_{2} A m m o n i u m c a r b o n a t e - ------------ \to A B$
$B a C O_{3} d i l H C l - --- \to B a C l_{2} P o t a s s i u m c h r o m a t e - ----------- \to B a C r O_{4} d i l H_{2} S O_{4} - ----- \to B a S O_{4} C D E$

So, $E$ is the Barium sulfate $(B a S O_{4})$

Suggest Corrections

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