Frank Brothers Solutions for ICSE Class 10 Chemistry Chapter 12 Practical Work

Practical Work

Solution 1:
Answer:
(a) Chloride ion: A small amount of the salt is taken in a test tube and conc. \( H_2SO_4 \) is added to it and then test tube is warmed, if a colourless gas with pungent odour is evolved then chloride ions are present in the salt. It can be confirmed by bringing a glass rod dipped in ammonia solution near the gas evolved, if dense white fumes are formed then presence of chloride ions is confirmed.

(b) It is a greenish yellow coloured gas with sharp pungent smell. It turns moist blue litmus paper red and finally bleaches it.

(c) It is a colourless gas with a smell of burnt sulphur. It turns moist blue litmus paper red.

(d) A small quantity of salt is taken in a test tube, dilute sulphuric acid is added to it. If a brisk effervescence is observed, then the gas is passed through lime water, If carbonate ion is present in the salt, the solution turns milky.

(e) Zinc hydroxide is an amphoteric hydroxide. It can be identified by treating the salt solution with sodium hydroxide, if a white ppt. is formed and if it gets dissolved in excess of sodium hydroxide then it may be zinc hydroxide.
In simple words: These are laboratory tests to identify different types of ions in salts - we use specific chemicals and observe changes like gas formation, color changes, or precipitation to confirm which ions are present.

📝 Teacher's Note: Demonstrate these tests with actual samples in the lab. Students remember better when they see the color changes and smell the gases. Always emphasize safety protocols when handling acids.

🎯 Exam Tip: Write the exact color of gases and precipitates formed - examiners look for specific observations like "dense white fumes" or "greenish yellow gas".

Solution 2:
Answer:
(a) Copper carbonate
(b) Selenium
(c) Nitre - a nitrate of potassium
(d) Anhydrous calcium sulphate
In simple words: These are chemical names for common substances - copper carbonate is a blue-green compound, selenium is a non-metal element, nitre is saltpeter used in fertilizers, and anhydrous calcium sulphate is dried gypsum.

📝 Teacher's Note: Connect these to everyday examples - show students actual samples of copper carbonate (green patina on copper), discuss selenium in supplements, and calcium sulphate in plaster of Paris.

🎯 Exam Tip: Remember the common names - nitre for potassium nitrate, and anhydrous means "without water" for calcium sulphate.

Solution 3:
Answer:
(a) \( NH_3 \): It has a strong pungent smell and turns moist red litmus blue. It can also be tested by bringing a glass rod dipped in conc. HCl in contact with the gas, if the gas is ammonia then it will produce dense white fumes near the glass rod.

(b) Oxygen: It is an odourless and colourless gas and turns alkaline pyrogallol brown. It can also be tested by bringing a lighted splinter near the gas, if the splinter starts glowing, the gas is oxygen.

(c) Water vapour: These are colourless vapours and are neutral to litmus. These can be tested by first condensing on the walls (cooler parts) of the test tube and then by adding anhydrous copper sulphate to the collected liquid, if the white colour of copper sulphate changes to blue, then the gas is water vapour.
In simple words: These are simple tests to identify common gases - ammonia smells strong and makes litmus blue, oxygen makes things burn brighter, and water vapor turns white copper sulphate blue when it condenses.

📝 Teacher's Note: Use the glowing splint test for oxygen as a memorable demonstration. Students love seeing the splint reignite. For ammonia, the white fumes test is very visual and convincing.

🎯 Exam Tip: Always mention both tests for each gas - the simple observation (smell, color) and the confirmatory chemical test for full marks.

Solution 4:
Answer:
Flame test:
1. Make a loop at the tip of the platinum wire and dip it in conc. HCl.
2. Put it on the non luminous part of the flame, to see if it gives colour. Repeat the process till it gives no colour to the flame.
3. Prepare a paste of the given salt in a watch glass using conc. HCl.
4. Load the loop of the wire with this prepared paste and introduce it into the non luminous flame of the bunsen burner and then observe the colour of the flame indicating different elements.
In simple words: Flame test is like a fingerprint for metals - each metal gives its own special color when heated in a flame, helping us identify which metal is present in the salt.

📝 Teacher's Note: Emphasize the importance of cleaning the wire between tests. Show students the characteristic colors - sodium gives yellow, potassium gives violet, copper gives green. Use colored filters to see potassium violet clearly.

🎯 Exam Tip: Write the complete procedure step by step. Mention "non-luminous flame" and "cleaning the wire" - these are key points examiners look for.

Solution 5:
Answer:
(a) When \( NH_3 \) solution is added to \( CuSO_4 \) solution drop by drop, ammonium ions completely react with copper sulphate and precipitates of copper hydroxide are formed

\( CuSO_4(aq) + 2NH_3(aq) + 2H_2O(l) \rightarrow Cu(OH)_2(s) + (NH_4)_2SO_4(aq) \)

But when we add excess of ammonia, the precipitate dissolves and a soluble complex is formed.

\( CuSO_4(aq) + 4NH_3(aq) \rightarrow [Cu(NH_3)_4]SO_4(aq) \)

(b) When Caustic soda solution is added to \( Cu(NO_3)_2 \) solution, the following reaction takes place:

\( Cu(NO_3)_2(aq) + 2 NaOH(aq) \rightarrow Cu(OH)_2(s) + 2 NaNO_3(aq) \)

When we heat it further, the greenish blue copper hydroxide decomposes to form slightly black ppt. of copper oxide.

\( Cu(OH)_2(s) \xrightarrow{\Delta} CuO(s) + H_2O(g) \)

(c) Common salt solution is added to silver nitrate solution.

\( AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq) \)

Common salt is NaCl, in aqueous medium it ionizes to form \( Na^+ \) and \( Cl^- \) ions.

\( NaCl(aq) \rightarrow Na^+(aq) + Cl^-(aq) \)
\( Cl^-(aq) + Ag^+(aq) \rightarrow AgCl(s) \)

On adding ammonia solution to it, the silver chloride gets dissolved and the following reaction occurs:

\( AgCl(s) + 2NH_3(aq) \rightarrow [Ag(NH_3)_2]^+(aq) + Cl^-(aq) \)

(d) When Lead nitrate solution is treated with calcium chloride solution it forms white ppt. of lead chloride and calcium nitrate

\( CaCl_2(s) \rightarrow Ca^{2+}(aq) + Cl^-(aq) \)

\( Pb(NO_3)_2(aq) \rightarrow Pb^{2+}(aq) + NO_3^-(aq) \)
\( Pb^{2+}(aq) + NO_3^-(aq) + Ca^{2+}(aq) + Cl^-(aq) \rightarrow PbCl_2 + Ca(NO_3)_2 \)

On further heating and cooling the products, calcium nitrate decomposes upon heating to release nitrogen dioxide:

\( 2 Ca(NO_3)_2 \rightarrow 2 CaO + 4 NO_2 + O_2 \)
In simple words: These reactions show how different chemicals interact - some form precipitates (solid particles), some dissolve to form complexes (special combinations), and some decompose when heated to form new substances.

📝 Teacher's Note: Use these reactions to teach the concept of precipitation, complex formation, and thermal decomposition. Show students the beautiful blue complex of copper with ammonia.

🎯 Exam Tip: Always write balanced chemical equations and mention the physical state (s), (l), (g), (aq). Describe the color and nature of precipitates formed.

Solution 7:
Answer:
Add concentrated hydrochloric acid to both the samples. Only \( MnO_2 \) releases greenish yellow chlorine gas.
In simple words: When you add strong acid to manganese dioxide, it produces chlorine gas which has a greenish-yellow color and sharp smell, but other substances don't react this way.

📝 Teacher's Note: Demonstrate this test in a fume hood due to chlorine gas. Explain that MnO₂ acts as an oxidizing agent. This is a classic test for identifying manganese dioxide.

🎯 Exam Tip: Mention the characteristic "greenish yellow" color of chlorine gas and that only MnO₂ gives this specific reaction with concentrated HCl.

Solution 8:
Answer:
C is silver chloride which is soluble in ammonia.
Pungent smelling gas B is ammonia.
White solid A is ammonium chloride

\( NH_4Cl(s) \xrightarrow{heat} NH_3(g) + HCO + NaCl \)
A                              B
\( NH_4(g) \xrightarrow{red litmus} blue litmus \)
\( NH_4Cl(s) \xrightarrow{AgNO_3, HNO_3} AgCl(s) \)
A                                                  C
In simple words: When ammonium chloride is heated, it breaks down into ammonia gas (which smells strong and turns red litmus blue) and can form silver chloride when tested with silver nitrate.

📝 Teacher's Note: This is sublimation of ammonium chloride. Heat the compound gently and show students how it directly converts to gas without melting. The litmus test for ammonia is a classic demonstration.

🎯 Exam Tip: Write the balanced equation for sublimation and mention that ammonia turns red litmus blue (alkaline nature). Identify each substance A, B, and C clearly.

Solution 9:
Answer:
(a) Distilled water: same as that of universal indicator
(b) Acid rain: red
(c) Soap solution: purple
(d) soil containing slaked lime: green
(e) Gastric juices: orange
In simple words: Universal indicator shows different colors for different pH levels - red for strong acids, orange for weak acids, green for neutral/slightly basic, and purple for strong bases.

📝 Teacher's Note: Use the universal indicator color chart to show students the pH scale. Let them test different household substances like lemon juice, baking soda solution, and soap water.

🎯 Exam Tip: Remember the color sequence: red (acidic) → orange → yellow → green (neutral) → blue → purple (basic). Match the substance's nature with the appropriate color.

Solution 10:
Answer:
(a) Ammonium sulphate on reacting with sodium hydroxide liberates ammonia gas:
\( (NH_4)_2 SO_4 (aq) + 2NaOH (l) \rightarrow Na_2SO_4 (aq) + 2NH_3 (l) + 2H_2O (l) \)
Sodium sulphate on reacting with sodium hydroxide undergoes the following change:
\( NaOH + Na_2SO_4 \rightarrow NaOH + Na_2SO_4 \)
However the products would be the same as the reactants, therefore providing no observation for a visible chemical reaction.

(b) Sodium hydroxide reacts with zinc nitrate and zinc hydroxide is formed which is soluble in excess of sodium hydroxide
\( Zn(NO_3)_2(aq) + 2NaOH(aq) \rightarrow Zn(OH)_2(s) + 2NaNO_3(aq) \)
Sodium hydroxide reacts with calcium nitrate and forms white ppt. of calcium hydroxide which is insoluble in excess of sodium hydroxide.
\( Ca(NO_3)_2 + 2NaOH \rightarrow Ca(OH)_2 + 2NaNO_3 \)

(c) Iron(III) chloride on reacting with sodium hydroxide forms brown ppt.
\( FeCl_3(aq) \)      +      \( 3NaOH(aq) \)    →      \( Fe(OH)_3(s) \)      +      \( 3NaCl(aq) \)

brown

Iron (II) chloride on reacting with sodium hydroxide forms green ppt.

\( FeCl_2(aq) \)      +      \( 2NaOH(aq) \)    →      \( Fe(OH)_2(s) \)      +      \( 2NaCl(aq) \)

green
In simple words: Different salts react differently with sodium hydroxide - some release gases (like ammonia), some form precipitates of different colors (brown, green, white), and some dissolve in excess base while others don't.

📝 Teacher's Note: Emphasize the difference between Fe(II) and Fe(III) compounds by their precipitate colors. This is a common way to distinguish between different oxidation states of iron.

🎯 Exam Tip: Always specify the color of precipitates formed - "brown ppt." for Fe(OH)₃ and "green ppt." for Fe(OH)₂. Mention solubility in excess NaOH for zinc hydroxide.

Solution 1999-1:
Answer:
(a) Sodium chloride solution and sodium nitrate solution can be distinguished by using conc. Sulphuric acid. To the salt solution, add freshly prepared ferrous sulphate solution and pour a few drops of conc. \( H_2SO_4 \) along the sides of the tube. If it's sodium nitrate solution then a brown ring would appear at the junction of the two liquid layers. But if its sodium chloride solution, it would not undergo any visible reaction.

(b) Sodium sulphate solution and sodium chloride solution can be distinguished by using barium chloride solution. Barium chloride solution on being added to sodium sulphate solution forms a white precipitate which is insoluble in conc. HCl. Whereas sodium chloride shows no reaction with barium chloride solution.

(c) Calcium nitrate solution and zinc nitrate solution can be distinguished by sodium hydroxide. Sodium hydroxide reacts with zinc nitrate and zinc hydroxide is formed which is soluble in excess of sodium hydroxide.

\( Zn(NO_3)_2(aq) + 2NaOH(aq) \rightarrow Zn(OH)_2(s) + 2NaNO_3(aq) \)
Sodium hydroxide reacts with calcium nitrate and forms white ppt. of calcium hydroxide insoluble in excess of sodium hydroxide.
\( Ca(NO_3)_2 + 2NaOH \rightarrow Ca(OH)_2 + 2NaNO_3 \)
In simple words: These are specific chemical tests to tell apart similar-looking solutions - we use different reagents that react differently with each solution, showing unique colors, precipitates, or gas formation patterns.

📝 Teacher's Note: The brown ring test for nitrates is a classic qualitative analysis technique. Demonstrate each test separately and emphasize the specific observations that distinguish each pair of compounds.

🎯 Exam Tip: For each pair, clearly state which reagent to use and what specific observation confirms the identity. Write "brown ring" for nitrate test, "white precipitate insoluble in HCl" for sulphate test.

Solution 2000-1

Question. Calcium nitrate:
(i) with sodium hydroxide:
• \( Ca(NO_3)_2 \xrightarrow{NaOH} Ca(OH)_2 + NaNO_3 \)
• White curdy ppt. is insoluble in excess of NaOH.
(ii) with ammonium hydroxide:
• \( Ca(NO_3)_2 \xrightarrow{NH_4OH} Ca(OH)_2 + NH_4NO_3 \)
• No precipitation of \( Ca(OH)_2 \) even in excess of ammonium hydroxide.
Answer: Calcium nitrate reacts with sodium hydroxide to form calcium hydroxide (white curdy precipitate) and sodium nitrate. The calcium hydroxide precipitate does not dissolve in excess sodium hydroxide. With ammonium hydroxide, calcium hydroxide forms but no visible precipitate appears even with excess ammonium hydroxide.

📝 Teacher's Note: Demonstrate the difference in solubility behavior by showing students how calcium hydroxide precipitate remains visible with NaOH but disappears with NH₄OH. This helps students understand amphoteric behavior concepts.

🎯 Exam Tip: Always mention the color of precipitates and their behavior in excess reagent - this distinction between NaOH and NH₄OH reactions is frequently tested.

Question. Zinc nitrate:
(i) with sodium hydroxide:
• \( Zn(NO_3)_2 + 2NaOH \rightarrow Zn(OH)_2 + 2NaNO_3 \)
• On addition of excess of NaOH, white ppt. of \( Zn(OH)_2 \) dissolves.
(ii) with ammonium hydroxide:
• \( Zn(NO_3)_2 + 2NH_4OH \rightarrow Zn(OH)_2 + 2NH_4NO_3 \)
• The gelatinous white insoluble ppt. of \( Zn(OH)_2 \) dissolves in excess of ammonium hydroxide.
Answer: Zinc nitrate forms white gelatinous zinc hydroxide precipitate with both sodium hydroxide and ammonium hydroxide. The key difference is that this precipitate dissolves in excess of both reagents, showing zinc hydroxide's amphoteric nature.

📝 Teacher's Note: Emphasize that zinc hydroxide is amphoteric - it dissolves in excess of both acids and bases. Use this as a perfect example to teach amphoteric behavior.

🎯 Exam Tip: Remember "zinc dissolves in excess" - this amphoteric property distinguishes zinc from calcium and lead in qualitative analysis questions.

Question. Lead nitrate:
(i) with sodium hydroxide:
• \( Pb(NO_3)_2 + 2NaOH \rightarrow Pb(OH)_2 + 2NaNO_3 \)
• White curdy ppt. is soluble in excess of NaOH.
(ii) with ammonium hydroxide:
• \( Pb(NO_3)_2 + 2NH_4OH \rightarrow Pb(OH)_2 + 2NH_4NO_3 \)
• White curdy ppt. is insoluble in excess of ammonium hydroxide.
Answer: Lead nitrate reacts with sodium hydroxide to form white curdy lead hydroxide precipitate that dissolves in excess sodium hydroxide. With ammonium hydroxide, the same white precipitate forms but remains insoluble even in excess ammonium hydroxide.

📝 Teacher's Note: Help students remember that lead hydroxide behaves opposite to calcium hydroxide - it dissolves in excess NaOH but not in excess NH₄OH. Create a comparison chart for better understanding.

🎯 Exam Tip: Lead hydroxide dissolves only in excess NaOH, not NH₄OH - this opposite behavior from calcium is a common exam trap question.

Solution 2001-1

Question. Answer:
\( Fe_2(SO_4)_3 + 6NH_4OH \rightarrow 2Fe(OH)_3 \downarrow + 3(NH_4)_2SO_4 \)
reddish brown
Reddish brown ppt. of ferric hydroxide is formed which is insoluble in excess of ammonium hydroxide.
Answer: When ferric sulfate reacts with ammonium hydroxide, it forms reddish brown ferric hydroxide precipitate and ammonium sulfate. The precipitate does not dissolve in excess ammonium hydroxide.

📝 Teacher's Note: Show students the distinctive reddish brown color of ferric hydroxide - this color is diagnostic for Fe³⁺ ions and helps in qualitative analysis of iron compounds.

🎯 Exam Tip: Always write the color "reddish brown" for Fe(OH)₃ precipitate - this specific color description earns full marks in practical exams.

Solution 2001-2

Question. On heating \( CuCO_3 \), CuO is formed which is black in colour.
Answer: When copper carbonate is heated, it decomposes to form black copper oxide and carbon dioxide gas is evolved.

📝 Teacher's Note: Demonstrate this reaction in class as the color change from green copper carbonate to black copper oxide is very striking and helps students remember thermal decomposition reactions.

🎯 Exam Tip: Always mention the color change and gas evolved in thermal decomposition reactions - these observation details are crucial for full marks.

Solution 2002-1

Question. Distinction between Zinc ion and Lead ion.
(i) \( ZnSO_4 + 2NH_4OH \rightarrow Zn(OH)_2 \downarrow + (NH_4)_2SO_4 \)
\( Zn(OH)_2 \) forms white gelatinous precipitate. In the presence of excess of ammonium hydroxide these precipitates get dissolved.
(ii) \( Pb(NO_3)_2 + 2NH_4OH \rightarrow Pb(OH)_2 \downarrow + 2NH_4NO_3 \)
\( Pb(OH)_2 \) forms white precipitate. This precipitate is insoluble in the presence of excess of ammonium hydroxide.
Answer: Both zinc and lead ions form white precipitates with ammonium hydroxide. The key difference is that zinc hydroxide dissolves in excess ammonium hydroxide while lead hydroxide remains insoluble in excess ammonium hydroxide.

📝 Teacher's Note: Use this table to help students memorize the color patterns after heating carbonates - zinc stays white, lead turns red, copper turns black. These are standard qualitative analysis results.

🎯 Exam Tip: Remember the simple pattern for carbonate heating: "White Zinc, Red Lead, Black Copper" - this mnemonic helps in both theory and practical exams.

Solution 2003-1

Question. When NaOH is added drop-wise to a solution of zinc sulphate, then the following reaction takes place:
\( ZnSO_4 + 2NaOH \rightarrow Zn(OH)_2 \downarrow + Na_2SO_4 \)
White gelatinous ppt.
When NaOH becomes in excess, then the following reaction takes place:
\( Zn(OH)_2 + 2NaOH \rightarrow Na_2ZnO_2 + 2H_2O \)
Colourless
Answer: Initially, zinc sulfate reacts with sodium hydroxide to form white gelatinous zinc hydroxide precipitate. When excess sodium hydroxide is added, the precipitate dissolves to form colorless sodium zincate solution.

📝 Teacher's Note: Demonstrate this reaction step by step - first the formation of white precipitate, then its dissolution in excess NaOH. This clearly shows amphoteric behavior in action.

🎯 Exam Tip: Write both equations clearly showing the two-step process - precipitate formation first, then dissolution in excess. This demonstrates understanding of amphoteric behavior.

Question. Dropwise addition of \( NH_4OH \):
\( CuSO_4 + 2NH_4OH \rightarrow Cu(OH)_2 + (NH_4)_2SO_4 \)
Blue ppt.
With excess of \( NH_4OH \), the precipitate of copper(II) hydroxide dissolves as:
\( Cu(OH)_2 + (NH_4)_2SO_4 + 2NH_4OH \rightarrow [Cu(NH_3)_4]SO_4 + 4H_2O \)
(In excess) Tetramine copper (II) Sulphate
Answer: Copper sulfate first forms blue copper hydroxide precipitate with ammonium hydroxide. In excess ammonium hydroxide, the precipitate dissolves to form deep blue tetramine copper(II) sulfate complex.

📝 Teacher's Note: Show students the beautiful color change from light blue precipitate to deep blue solution - this complex formation is a classic example of coordination chemistry in action.

🎯 Exam Tip: Mention the deep blue color of the copper-ammonia complex - this distinctive color change is a key identification test for copper ions.

Question. Silver chloride is formed on adding hydrochloric acid in silver nitrate solution. On adding excess of ammonium chloride, white ppt. gets dissolved.
\( 2AgCl(s) + 2NH_4OH(aq) \rightarrow Ag_2O(s) + H_2O(l) + 2NH_4Cl(aq) \)
Chlorine gas turns moist starch iodide paper black.
Purple colour of potassium permanganate gets discharged.
Brick effervescence is observed and colourless and odourless \( CO_2 \) gas is evolved in both the cases.
Answer: Silver chloride precipitate dissolves in excess ammonium hydroxide forming silver oxide. Chlorine gas can be identified by its ability to turn moist starch-iodide paper black and discharge the purple color of potassium permanganate. Carbon dioxide gas shows effervescence and is colorless and odorless.

📝 Teacher's Note: These are standard qualitative tests - emphasize the importance of using multiple confirmatory tests for gas identification, especially the color changes with indicator papers.

🎯 Exam Tip: Always mention both the starch-iodide test AND permanganate test for chlorine gas - using multiple tests shows thorough understanding of qualitative analysis.

Solution 2004-1

📝 Teacher's Note: Use this table as a reference chart for students to memorize the behavior patterns of different metal ions with sodium hydroxide - it's essential for qualitative analysis.

🎯 Exam Tip: Learn the solubility patterns: Zinc and Lead dissolve in excess NaOH (amphoteric), while Copper, Calcium, and Iron remain insoluble (basic hydroxides only).

Solution 2004-2

📝 Teacher's Note: Teach students to identify gases by their distinctive odors - H₂S smells like rotten eggs, Cl₂ is pungent, while CO₂ and H₂ are odorless. This sensory identification is crucial in practical work.

🎯 Exam Tip: Always mention the odor characteristics of gases in identification questions - it's often the easiest way to distinguish between similar-looking gases.

Solution 2005-1

Question. (i) Magnesium sulphate, Iron(II) sulphate
(ii) Zinc chloride, Iron(III) chloride
(iii) Lead nitrate
(iv) Copper nitrate
Answer: These are examples of different metal salts that can be identified through their characteristic reactions with various reagents. Each salt shows specific precipitation patterns and color changes when tested with standard qualitative analysis reagents like sodium hydroxide and ammonium hydroxide.

📝 Teacher's Note: Group these salts by their metal ions and teach students the systematic approach to salt identification - start with hydroxide tests, then move to specific confirmatory tests for each metal.

🎯 Exam Tip: Practice identifying these common salts by their hydroxide precipitation colors and solubility behaviors - this forms the foundation of qualitative analysis.

Solution 2006-1

Question. (a) 1. Calcium: white
2. Zinc: white
3. Lead: orangish red
4. Copper: black
Answer: These are the characteristic colors of metal oxides or hydroxides formed during various chemical tests. Calcium and zinc compounds typically appear white, lead compounds show orangish red color, and copper compounds appear black.

📝 Teacher's Note: Create color charts for students showing the characteristic colors of different metal compounds - visual memory aids are extremely helpful for qualitative analysis.

🎯 Exam Tip: Memorize these standard colors as they frequently appear in both theory and practical examinations for metal identification.

Question. (b) 1. Zinc nitrate and calcium nitrate solution can be distinguished by reaction with ammonium hydroxide. Zinc forms a white gelatinous ppt. whereas there is no precipitation of calcium hydroxide even with excess of ammonium hydroxide.
\( Zn(NO_3)_2 + 2NH_4OH \rightarrow Zn(OH)_2 \downarrow + 2NH_4NO_3 \)
White gelatinous ppt.
Answer: Zinc nitrate forms a white gelatinous precipitate with ammonium hydroxide while calcium nitrate shows no precipitation. This difference in behavior allows easy distinction between these two salts.

📝 Teacher's Note: Demonstrate this test practically - students can clearly see the precipitate formation with zinc while calcium solution remains clear, making this an excellent diagnostic test.

🎯 Exam Tip: When distinguishing between zinc and calcium salts, always use NH₄OH test - it's the most reliable and straightforward method.

Question. 1. (a) Sodium chloride solution and sodium nitrate solution can be distinguished by using conc. Sulphuric acid. To the salt solution, add freshly prepared ferrous sulphate solution and pour a few drops of conc. \( H_2SO_4 \) along the sides of the tube. If it's sodium nitrate solution then a brown ring would appear at the junction of the two liquid layers. But if its sodium chloride solution, it would not undergo any visible reaction.
Answer: The brown ring test is used to distinguish sodium nitrate from sodium chloride. With ferrous sulfate and concentrated sulfuric acid, sodium nitrate produces a brown ring at the interface while sodium chloride shows no reaction.

📝 Teacher's Note: This is the classic brown ring test for nitrates - emphasize the importance of careful layering of concentrated sulfuric acid and watching for the brown ring formation.

🎯 Exam Tip: Always mention "brown ring at the junction" for nitrate test - this specific location detail is crucial for full marks.

Question. 2. Iron (III) chloride solution and copper chloride solution can be distinguished by using ammonium hydroxide. Copper forms a blue ppt. of \( Cu(OH)_2 \) which is soluble in excess of ammonium hydroxide.
\( CuCl_2 + 2NH_4OH \rightarrow Cu(OH)_2 + 2NH_4Cl \)
Whereas iron(III) forms a reddish brown ppt. of \( Fe(OH)_3 \) which is insoluble even in excess of ammonium hydroxide.
\( FeCl_3 + 3NH_4OH \rightarrow Fe(OH)_3 + 3NH_4Cl \)
Answer: Iron(III) chloride and copper chloride can be distinguished by their different colored precipitates with ammonium hydroxide. Copper forms a blue precipitate that dissolves in excess while iron(III) forms a reddish brown precipitate that remains insoluble in excess.

📝 Teacher's Note: Show students both precipitates side by side - the blue vs reddish brown color difference is striking and the solubility difference in excess NH₄OH is also clearly visible.

🎯 Exam Tip: Remember "blue copper dissolves, brown iron doesn't" - this simple rule helps distinguish these two important metal chlorides.

Solution 2006-2:

Question. Match the following:

Answer: 1-C, 2-D, 3-E, 4-A, 5-B
In simple words: Each substance has unique chemical properties that help identify it through simple tests - like color changes, gas formation, or reactions with common chemicals.

📝 Teacher's Note: Use actual demonstrations of these tests in class - students remember practical observations much better than theoretical descriptions. Keep samples of starch iodide paper and common chemicals for quick demos.

🎯 Exam Tip: Remember the key identifying features: chlorine turns starch iodide paper blue, copper nitrate gives brown gas with acids, ferrous sulphate gives green precipitate, ammonium salts release ammonia, lead carbonate turns yellow when heated.

Solution 2007-1:

📝 Teacher's Note: This appears to be a reference table for salt analysis. Teach students to memorize common anions and their charges as this forms the foundation for qualitative analysis.

🎯 Exam Tip: Practice writing chemical formulas with correct charges - examiners often deduct marks for incorrect ionic charges even if the identification is correct.

Solution 2008-1:

Answer: Iron(II) sulphate
In simple words: This is the chemical name for the compound also known as ferrous sulphate, which is an important iron salt used in chemistry experiments and medicine.

📝 Teacher's Note: Emphasize the difference between Iron(II) and Iron(III) compounds - the Roman numeral indicates the oxidation state of iron, which affects the compound's properties and color.

🎯 Exam Tip: Always write Iron(II) instead of ferrous and Iron(III) instead of ferric in IUPAC naming - modern chemistry exams prefer systematic nomenclature.