
A white coloured salt forms white sublimate in dry heating test and also gives ammonia on heating with caustic soda solution. Which of the following tests will be shown positive by the salt?
A.It will give greenish yellow gas on reaction with conc.${H_2}S{O_4}$.
B.It will give red gas by heating with ${K_2}C{r_2}{O_7}$ and conc.${H_2}S{O_4}$.
C.It will give colourless gas with dil. ${H_2}S{O_4}$
D.It will give a ring test.
Answer
451.2k+ views
Hint: We have to know that qualitative analysis is about analysis of cation and anion present in the given compound. We have to know that physical procedures such as observing color of substance, smell of substance (or) taste of substance contains only limited scope. So, besides physical examination of the compound, chemical examination is also done.
Complete answer:
In the question, it is given that white sublimate is formed when the inorganic salt undergoes a dry testing tube. We have to know that while performing the test, a little quantity of salt is heated in the dry test tube. Certain salts undergo decomposition, when the salt is heated. When the salt undergoes decomposition, we can observe evolution of gas (or) some characteristics change in residue color.
The formation of white sublimates in dry heating tests predicts the presence of ammonium ions in the compound. So, the salt which is white colored should be ammonium chloride while undergoing dry heating test forms white sublimate. Ammonia gas is liberated when ammonium chloride is heated with a solution of caustic soda. We can write the chemical equation as,
$N{H_4}Cl + NaOH \to NaCl + N{H_3} \uparrow + {H_2}O$
When we heat ammonium chloride with potassium dichromate and concentrated hydrochloric acid, a red gas is liberated out. Red gas indicates the presence of chromyl chloride. We can write the chemical equation as,
$4N{H_4}Cl + {K_2}C{r_2}{O_7} + 3{H_2}S{O_4} \to {K_2}S{O_4} + 2{\left( {N{H_4}} \right)_2}S{O_4} + 2Cr{O_2}C{l_2} \uparrow + 3{H_2}O$
From this we can conclude that red gas is formed by heating with ${K_2}C{r_2}{O_7}$ and conc.${H_2}S{O_4}$.
Option (B) is correct.
Note:
We have to know that other systematic methods for qualitative analysis of an inorganic salt are,
Charcoal cavity test
Cobalt nitrate test
Flame test
Borax bead test
Concentrated sulfuric acid
Complete answer:
In the question, it is given that white sublimate is formed when the inorganic salt undergoes a dry testing tube. We have to know that while performing the test, a little quantity of salt is heated in the dry test tube. Certain salts undergo decomposition, when the salt is heated. When the salt undergoes decomposition, we can observe evolution of gas (or) some characteristics change in residue color.
The formation of white sublimates in dry heating tests predicts the presence of ammonium ions in the compound. So, the salt which is white colored should be ammonium chloride while undergoing dry heating test forms white sublimate. Ammonia gas is liberated when ammonium chloride is heated with a solution of caustic soda. We can write the chemical equation as,
$N{H_4}Cl + NaOH \to NaCl + N{H_3} \uparrow + {H_2}O$
When we heat ammonium chloride with potassium dichromate and concentrated hydrochloric acid, a red gas is liberated out. Red gas indicates the presence of chromyl chloride. We can write the chemical equation as,
$4N{H_4}Cl + {K_2}C{r_2}{O_7} + 3{H_2}S{O_4} \to {K_2}S{O_4} + 2{\left( {N{H_4}} \right)_2}S{O_4} + 2Cr{O_2}C{l_2} \uparrow + 3{H_2}O$
From this we can conclude that red gas is formed by heating with ${K_2}C{r_2}{O_7}$ and conc.${H_2}S{O_4}$.
Option (B) is correct.
Note:
We have to know that other systematic methods for qualitative analysis of an inorganic salt are,
Charcoal cavity test
Cobalt nitrate test
Flame test
Borax bead test
Concentrated sulfuric acid
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