Practical Chemistry

Questions

Question 1(2006)

From the list of substances given — Ammonium sulphate, Lead carbonate, Chlorine, Copper nitrate, Ferrous sulphate — State : A substance that turns moist starch iodide paper blue.

Answer

Chlorine

Cl₂ + 2KI ⟶ 2KCl + I₂
[Starch + I₂ ⟶ blue black colour]

Question 2(2006)

State what is observed when excess of ammonia passed through an aqueous solution of lead nitrate.

Answer

Chalky white precipitate of lead hydroxide is formed which is insoluble in excess of ammonia.

Pb(NO₃)₂ + 2NH₄OH ⟶ 2NH₄NO₃ + Pb(OH)₂ ↓ [Chalky white ppt. formed]

Question 3(2006)

Give one test each to distinguish between the following pairs of chemical solutions

(i) Zn(NO₃)₂ and Ca(NO₃)₂

(ii) NaNO₃ and NaCl

(iii) Iron [III] chloride and copper chloride .

Answer

(i) When NaOH is added to the given soln., Zn(NO₃)₂ reacts to form a gelatinous white ppt. which dissolves in excess of NaOH whereas, Ca(NO₃)₂ forms a milky white ppt. which is insoluble in excess of NaOH. Hence, the two can be distinguished.

(ii) Add conc. sulphuric acid to the given salts, NaCl will react to produce HCl gas, which produces dense white fumes when a glass rod dipped in ammonia soln. is brought near it. If we add copper turnings along with conc. sulphuric acid, NaNO₃ will react to give reddish brown nitrogen dioxide. Hence, the two salts can be distinguished.

(iii) When we add NaOH solution to the given solutions, Iron [III] chloride forms a reddish brown ppt. of Fe(OH)₃ whereas, copper chloride gives a pale blue ppt of Cu(OH)₂.

Question 4(2006)

Give a reason why carbon dioxide and sulphur dioxide cannot be distinguished by using lime water.

Answer

Both, carbon dioxide and sulphur dioxide turns lime water milky due to the formation of CaCO₃ and CaSO₃, respectively. Hence, lime water cannot be used to distinguish between the two.

Ca(OH)₂ + CO₂ ⟶ CaCO₃ ↓ [white ppt. - insoluble] + H₂O

Ca(OH)₂ + SO₂ ⟶ CaSO₃ ↓ [white ppt. - insoluble] + H₂O

Question 1(2007)

Salts A, B, C, D and E undergo reactions (i) to (v) respectively. Identify the anion present in each salt.

(i) When AgNO₃ solution is added to a soln. of A, a white precipitate, insoluble in dilute nitric acid, is formed.

(ii) Addition of dil. HCl to B produces a gas which turns lead acetate paper black.

(iii) When a freshly prepared solution of FeSO₄ is added to a soln. of C and conc. H₂SO₄ is gently poured from the side of the test-tube, a brown ring is formed.

(iv) When dil. H₂SO₄ is added to D a gas is produced which turns acidified K₂Cr₂O₇ soln. from orange to green.

(v) Addition of dil. HCl to E produced an effervescence. The gas produced turns limewater milky but does not effect acidified K₂Cr₂O₇ soln.

Answer

(i) Chloride (Cl^1-)

(ii) Sulphide (S^2-)

(iii) Nitrate (NO₃^1-)

(iv) Sulphite (SO₃^2-)

(v) Carbonate (CO₃^2-)

Question 2(2007)

How will the addition of barium chloride soln. help to distinguish between dil. HCl and dil. H₂SO₄.

Answer

Barium chloride reacts with dil H₂SO₄ to form a white ppt. of barium sulphate.

BaCl₂ + H₂SO₄ [dil.] ⟶ 2HCl + BaSO₄ ↓ [white ppt.]

Whereas, dil. HCl does not form a white ppt. Hence, the two can be distinguished.

BaCl₂ + HCl [dil.] ⟶ No white ppt.

Question 1(2008)

The salt which in solution gives a pale green ppt. with NaOH soln. and a white ppt. with barium chloride solution is :

1. Iron (III) sulphate
2. Iron (II) sulphate
3. Iron (II) chloride
4. Iron (III) chloride

Answer

Iron (II) sulphate

Pale green ppt. of Fe(OH)₂ and white ppt. of barium sulphate are formed.

Question 1(2009)

CO₂ and SO₂ gas can be distinguished using :

1. moist blue litmus paper
2. lime water
3. acidified K₂Cr₂O₇ paper
4. none of the above.

Answer

acidified K₂Cr₂O₇ (potassium dichromate) paper

Reason

There is no effect of CO₂ gas on potassium dichromate whereas SO₂ turns acidified potassium dichromate from orange to clear green.

K₂Cr₂O₇ + H₂SO₄ + 3SO₂ ⟶ K₂SO₄ + Cr₂(SO₄)₃ + H₂O

Question 2(2009)

Identity the substance 'R' based on the information given below : The pale green solid 'R' turns reddish brown on heating. It's aqueous solution gives a white precipitate with barium chloride solution. The precipitate is insoluble in mineral acids.

Answer

The substance 'R' is Ferrous sulphate (FeSO₄.7H₂O).

Reason

When heated strongly, the hydrous pale green ferrous sulphate (FeSO₄.7H₂O), loses it's water of crystallization and decomposes to form brown ferric oxide Fe₂O₃ along with sulphur dioxide (SO₂) and sulphur trioxide (SO₃).

FeSO₄.7H₂O $\xrightarrow{\Delta}$ FeSO₄ + 7H₂O

2FeSO₄ $\xrightarrow{\Delta}$ Fe₂O₃ + SO₂ + SO₃

Ferrous sulphate on reaction with barium chloride soln., forms a white ppt. of barium sulphate (BaSO₄) which is insoluble in mineral acids.

Question 3(2009)

Give one test each to distinguish between the following pairs of chemical solutions:

(i) ZnSO₄ and ZnCl₂

(ii) FeCl₂ and FeCl₃

(iii) Calcium nitrate soln. and Calcium chloride soln.

Answer

(i) When BaCl₂ soln. is added to ZnSO₄, a white ppt. of BaSO₄ is formed, whereas, no ppt. is formed in case of ZnCl₂. Hence, the two solns. can be distinguished.

BaCl₂ + ZnSO₄ ⟶ BaSO₄ ↓ [white ppt.] + ZnCl₂

BaCl₂ + ZnCl₂ ⟶ No white ppt.

(ii) When NaOH soln. is added to the given solns., FeCl₂ reacts to form a dirty green ppt. of Fe(OH)₂ whereas, FeCl₃ reacts to form a reddish brown ppt. of Fe(OH)₃.

(iii) Add silver nitrate soln. to the given solns., calcium chloride reacts to form a white ppt. which is soluble in NH₄OH and insoluble in dil. HNO₃. The other soln. is calcium nitrate.

CaCl₂ + 2AgNO₃ ⟶ 2AgCl ↓ [white ppt.] + Ca(NO₃)₂

Ca(NO₃)₂ + AgNO₃ ⟶ no white ppt.

Question 1(2010)

Select the correct answer from A, B, C, D and E –

(A) : Nitroso Iron (II) sulphate

(B) : Iron (III) chloride

(C) : Chromium sulphate

(D) : Lead (II) chloride

(E) : Sodium chloride.

The compound which is responsible for the green colour formed when SO₂ is bubbled through acidified potassium dichromate solution.

Answer

Chromium sulphate

K₂Cr₂O₇ + H₂SO₄ + 3SO₂ ⟶ K₂SO₄ + Cr₂(SO₄)₃ + H₂O

Question 2(2010)

State your observation – (i) A piece of moist blue litmus paper (ii) paper soaked in potassium permanganate solution - is introduced in each case into a jar of sulphur dioxide.

Answer

(i) Moist blue litmus turns red.

(ii) Sulphur dioxide turns potassium permanganate from pink to clear colourless.

2KMnO₄ + 2H₂O + 5SO₂ ⟶ K₂SO₄ + 2MnSO₄ + 2H₂SO₄

Question 3(2010)

Write the equation for the reaction of magnesium sulphate solution with barium chloride solution.

Answer

MgSO₄ + BaCl₂ ⟶ BaSO₄ ↓ [white ppt.] + MgCl₂

Question 1 (2011)

Choose from the list of substances – Acetylene gas, aqua fortis, coke, brass, barium chloride, bronze, platinum.

An aqueous salt solution used for testing sulphate radical.

Answer

Barium chloride.

Question 1(2012)

Name the gas which turns acidified potassium dichromate clear green.

Answer

Sulphur dioxide gas (SO₂)

K₂Cr₂O₇ + H₂SO₄ + 3SO₂ ⟶ K₂SO₄ + Cr₂(SO₄)₃ + H₂O

Question 2(2012)

Identify the anion present in the following compounds :

(i) Compound X on heating with copper turnings and concentrated sulphuric acid liberates a reddish brown gas.

(ii) When a solution of compound Y is treated with silver nitrate solution a white precipitate is obtained which is soluble in excess of ammonium hydroxide solution.

(iii) Compound Z which on reacting with dilute sulphuric acid liberates a gas which turns lime water milky, but the gas has no effect on acidified potassium dichromate solution.

(iv) Compound L on reacting with Barium chloride solution gives a white precipitate insoluble in dilute hydrochloric acid or dilute nitric acid.

Answer

(i) Nitrate ion, NO₃^-

(ii) Chloride ion, Cl^-

(iii) Carbonate ion, CO₃^2-

(iv) Sulphate ion, SO₄^2-

Question 3(2012)

State one chemical test between each of the following pairs :

(i) Sodium carbonate and Sodium sulphite

(ii) Ferrous nitrate and Lead nitrate

(iii) Manganese dioxide and Copper (II) oxide

Answer

(i) When dil. sulphuric acid is added to sodium carbonate and heated, colourless, odourless gas is evolved which turns lime water milky and has no effect on KMnO₄ or K₂Cr₂O₇ solutions.

When dil. sulphuric acid is added to sodium sulphite and heated, colourless gas with suffocating odour is evolved which turns lime water milky. It turns acidified K₂Cr₂O₇ from orange to clear green and pink coloured KMnO₄ to clear colourless.

Hence, the two compounds can be distinguished.

(ii) When NaOH is added to each of the compounds, a dirty green precipitate of Iron [II] hydroxide [Fe(OH)₂] is formed in case of Ferrous nitrate whereas a chalky white precipitate of lead hydroxide [Pb(OH)₂] is formed in case of lead nitrate. Iron [II] Hydroxide [Fe(OH)₂] is insoluble in excess of NaOH, whereas Lead Hydroxide [Pb(OH)₂] is soluble in excess of NaOH. Hence, the two compounds can be distinguished.

(iii) When each of the compound is heated with conc. hydrochloric acid, greenish yellow (chlorine) gas is evolved in case of manganese dioxide and filtrate is brownish in colour whereas, no chlorine gas is evolved in case of copper (II) oxide and filtrate is bluish in colour.

MnO₂ + 4HCl ⟶ MnCl₂ + 2H₂O + Cl₂

CuO + 2HCl ⟶ CuCl₂ + H₂O

Question 4(2012)

State one observation : A zinc granule is added to copper sulphate solution.

Answer

Zinc displaces copper from copper sulphate and forms zinc sulphate soln. because zinc is more reactive than copper. Hence, the blue colour of copper sulphate solution changes to colourless.

Question 5(2012)

Give balanced equation for the reaction :

Silver nitrate solution and Sodium chloride solution.

Answer

AgNO₃ + NaCl ⟶ AgCl ↓ [white ppt.] + NaNO₃

Question 1(2013)

Give a chemical test to distinguish between :

(i) NaCl soln. and NaNO₃ soln.

(ii) HCl gas and H₂S gas.

(iii) Calcium nitrate soln. and zinc nitrate soln.

(iv) Carbon dioxide gas and sulphur dioxide gas.

Answer

(i) Add silver nitrate soln. to the given solns., sodium chloride reacts to form a white ppt. which is soluble in NH₄OH and insoluble in dil. HNO₃. The other soln. is sodium nitrate.

NaCl + AgNO₃ ⟶ AgCl ↓ [white ppt.] + NaNO₃

NaNO₃ + AgNO₃ ⟶ no white ppt.

(ii) Hydrogen sulphide gas turns moist lead acetate paper silvery black or black whereas, no change is observed in case of HCl gas.

Pb(CH₃COO)₂ [colourless] + H₂S ⟶ PbS [black] + 2CH₃COOH

(iii) When NaOH is added to the given soln., Zn(NO₃)₂ reacts to form a gelatinous white ppt. which dissolves in excess of NaOH whereas, Ca(NO₃)₂ forms a milky white ppt. which is insoluble in excess of NaOH. Hence, the two can be distinguished.

(iv) Carbon dioxide gas has no effect on acidified potassium permanganate (KMnO₄) and acidified potassium dichromate (K₂Cr₂O₇) solution whereas sulphur dioxide gas turns acidified potassium permanganate from pink to clear colourless and acidified potassium dichromate from orange to clear green.

2KMnO₄ + 2H₂O + 5SO₂ ⟶ K₂SO₄ + 2MnSO₄ + 2H₂SO₄

K₂Cr₂O₇ + H₂SO₄ + 3SO₂ ⟶ K₂SO₄ + Cr₂(SO₄)₃ + H₂O

Question 2(2013)

From

A : CO
B : CO₂
C : NO₂
D : SO₃

State which will not produce an acid on reaction with water.

Answer

CO
Reason — On reaction with water, CO₂ produces carbonic acid, NO₂ forms nitric acid and SO₃ forms sulphuric acid whereas CO forms carbon dioxide gas.

CO + H₂O ⟶ CO₂ + H₂

Question 1(2014)

Distinguish between : Sodium nitrate and sodium sulphite (using dilute sulphuric acid)

Answer

Sodium nitrate will not react with dilute sulphuric acid.

Sodium sulphite reacts with dil. sulphuric acid and colourless gas with suffocating odour is evolved which turns lime water milky.

Question 2(2014)

State your observation : When moist starch iodide paper is introduced into chlorine gas.

Answer

Chlorine gas turns moist starch iodide paper blue black.

Cl₂ + 2KI ⟶ 2KCl + I₂
[Starch + I₂ ⟶ blue black colour]

Question 3(2014)

The flame test with a salt P gave a brick red flame. What is the cation in P?

Answer

Cation in P is Ca²⁺ (calcium ion).

Question 4(2014)

Gas Q turns moist lead acetate paper silvery black. Identify Q. pH of R is 10. What kind of substance is R?

Answer

The gas Q is H₂S (Hydrogen sulphide).

Pb(CH₃COO)₂ [colourless] + H₂S ⟶ PbS [black] + 2CH₃COOH

R is an alkaline substance.

Question 1(2015)

Select the gas that has a characteristic rotten egg smell. [ammonia, ethane, hydrogen chloride, hydrogen sulphide, ethyne]

Answer

Hydrogen sulphide

Question 2(2015)

State one relevant observation : When hydrogen sulphide gas is passed through lead acetate solution.

Answer

A silvery black precipitate of PbS is formed.

Pb(CH₃COO)₂ [colourless] + H₂S ⟶ PbS [black] + 2CH₃COOH

Question 3(2015)

Identify the anion present in each of the following compounds: A, B, C :

(i) Salt 'A' reacts with conc. H₂SO₄ producing a gas which fumes in moist air and gives dense fumes with ammonia.

(ii) Salt 'B' reacts with dil. H₂SO₄ producing a gas which turns lime water milky but has no effect on acidified potassium dichromate solution.

(iii) When barium chloride solution is added to salt solution. 'C' a white precipitate insoluble in dilute hydrochloric acid is obtained.

Answer

(i) Chloride ion (Cl^–)

(ii) Carbonate ion (CO₃^2–)

(iii) Sulphate ion (SO₄^2–)

Question 4(2015)

Identify the cation present in each of the following compounds — W, X, Y, Z :

(i) To solution 'W', ammonium hydroxide is added in minimum quantity first and then in excess. A dirty white precipitate is formed which dissolves in excess to form a clear solution.

(ii) To solution 'X' ammonium hydroxide is added in minimum quantity first and then in excess. A pale blue precipitate is formed which dissolves in excess to form a clear inky blue solution.

(iii) To solution 'Y' a small amount of sodium hydroxide is added slowly and then in excess. A white insoluble precipitate is formed.

(iv) To salt 'Z' - Ca(OH)₂ soln. is added and heated. A pungent smelling gas turning moist red litmus paper blue is obtained.

Answer

(i) Zn²⁺ (Zinc Ion)

ZnSO₄ + 2NH₄OH ⟶ (NH₄)₂SO₄ + Zn(OH)₂ ↓

Zn(OH)₂ + (NH₄)₂SO₄ + 2NH₄OH [in excess] ⟶ 4H₂O + [Zn(NH₃)₄]SO₄

(ii) Cu²⁺ (Copper (II) Ion)

CuSO₄ + 2NH₄OH ⟶ (NH₄)₂SO₄ + Cu(OH)₂ ↓

Cu(OH)₂ + (NH₄)₂SO₄ + 2NH₄OH [in excess] ⟶ 4H₂O + [Cu(NH₃)₄]SO₄

(iii) Ca²⁺ (Calcium Ion)

CaCl₂ + 2NaOH ⟶ 2NaCl + Ca(OH)₂ ↓

(iv) NH₄⁺ (Ammonium Ion)

2NH₄Cl + Ca(OH)₂ ⟶ CaCl₂ + 2H₂O + 2NH₃

Question 1(2016)

Identify the cations in each of the following cases :

(i) NaOH solution when added to the solution 'A' - gives a reddish brown precipitate.

(ii) NH₄OH solution when added to the solution 'B' - gives a white ppt which does not dissolve in excess.

(iii) NaOH solution when added to the solution 'C' - gives a white ppt which is insoluble in excess.

Answer

(i) Ferric (Fe³⁺) ion

(ii) Lead (Pb²⁺) ion

(iii) Calcium (Ca²⁺) ion

Question 1(2017)

Choose the correct answer from the options – A chloride which forms a precipitate that is soluble in excess of ammonium hydroxide, is :

1. Calcium chloride
2. Ferrous chloride
3. Ferric chloride
4. Copper chloride.

Answer

Copper chloride.

Question 2(2017)

Identify the substance underlined – Cation that does not form a precipitate with ammonium hydroxide but forms one with sodium hydroxide.

Answer

Calcium [Ca²⁺] ions

Question 3(2017)

Identify the salts P and Q from the observations given below :

(i) On performing the flame test salt P produces a lilac coloured flame and it's solution gives a white precipitate with silver nitrate solution, which is soluble in ammonium hydroxide solution.

(ii) When dilute HCl is added to a salt Q, a brisk effervescence is produced and the gas turns lime water milky. When NH₄OH solution is added to the above mixture [after adding dilute HCl], it produces a white precipitate which is soluble in excess NH₄OH solution.

Answer

(i) P is potassium chloride.

Reason — K⁺ ions produces lilac coloured flame and Cl^- ions react with silver nitrate to form silver chloride ppt. (soluble in excess of ammonium hydroxide).

(ii) Q is zinc carbonate.

Reason — CO₃^2- ions produces carbon dioxide with HCl. Zinc chloride forms white ppt. with ammonium hydroxide ( soluble in excess of ammonium hydroxide).

Question 1(2018)

State one relevant observation - Barium chloride solution is slowly added to sodium sulphate solution.

Answer

A white ppt. is obtained which is insoluble in dil. HCl or dil. HNO₃

Na₂SO₄ + BaCl₂ ⟶ BaSO₄ ↓ [white ppt.] + 2NaCl

BaSO₄ is insoluble in dil. hydrochloric acid.

Question 2(2018)

Give a chemical test to distinguish between following pairs of chemicals :

(i) Lead nitrate solution and zinc nitrate solution.

(ii) Sodium chloride solution and sodium nitrate solution

Answer

(i) When NaOH is added to each of the compounds, lead nitrate forms a chalky white precipitate of lead hydroxide [Pb(OH)₂] whereas a gelatinous white precipitate of zinc hydroxide [Zn(OH)₂] is formed in case of zinc nitrate.

(ii) Add silver nitrate soln. to the given solns., sodium chloride reacts to form a white ppt. which is soluble in NH₄OH and insoluble in dil. HNO₃. The other soln. is sodium nitrate.

NaCl + AgNO₃ ⟶ AgCl + NaNO₃

NaNO₃ + AgNO₃ ⟶ no white ppt.

Question 1(2019)

State one observation for the following : Lead nitrate is heated strongly in a test tube.

Answer

White crystalline solid turns buff yellow on heating, gives crackling sound, melts and fuses with the glass. Nitrogen dioxide (reddish brown) and Oxygen (colourless) gases are evolved.

$\underset{\text{[lead nitrate (white)]}}{2\text{Pb(NO}_3)_2} \overset{\Delta}{\longrightarrow} \underset{\text{[litharge - buff yellow]}}{2\text{PbO}} + \underset{\text{[Nitrogen dioxide]}}{4\text{NO}_2} + \underset{\text{[Oxygen]}}{\text{O}_2}$

Question 2(2019)

Distinguish between the following pairs of compounds using the reagent given in the bracket.

(i) Manganese dioxide and copper [II] oxide [using conc. HCl]

(ii) Ferrous sulphate solution and ferric sulphate solution [using sodium hydroxide solution]

Answer

(i) When each of the compound is heated with conc. hydrochloric acid, greenish yellow (chlorine) gas is evolved in case of manganese dioxide and filtrate is brownish in colour whereas, no chlorine gas is evolved in case of copper (II) oxide and filtrate is bluish in colour.

MnO₂ + 4HCl ⟶ MnCl₂ + 2H₂O + Cl₂

CuO + 2HCl ⟶ CuCl₂ + H₂O

(ii) When sodium hydroxide is added to the two solns., ferrous sulphate solution gives a dirty green ppt. of Fe(OH)₂ whereas, ferric sulphate solution forms a reddish brown ppt. of Fe(OH)₃. Hence, the two compounds can be distinguished.

Question 1(2020)

State one relevant observation for the following reaction : Zinc carbonate is heated strongly.

Answer

The original white colour turns yellow on heating. Colourless, odourless carbon dioxide gas is evolved.

$\begin{matrix} \text{ZnCO}_3 &\longrightarrow & \text{ZnO} & + & \text{CO}_2 \\ \text{[Zinc} & & \text{[Zinc oxide]} & & \text{[Carbon} \\ \text{carbonate]} & & \text{[yellow - hot]} & & \text{dioxide]} \\ \text{[white]} & & \text{[white - cold]} \end{matrix}$

Question 2(2020)

State one relevant reason for the following : Hydrated copper sulphate crystals turn white on heating.

Answer

The blue coloured hydrous copper sulphate changes to white anhydrous copper sulphate as the water of crystallization is removed on heating.

CuSO₄.5H₂O ⟶ CuSO₄ + 5H₂O

Question 3(2020)

Match the gases given in column I to the identification of the gases mentioned in column II

Answer

Question 4(2020)

Distinguish between the following pairs of compounds using a reagent as a chemical test:

(i) Calcium nitrate and Zinc nitrate solution

(ii) Magnesium chloride and Magnesium nitrate solution

Answer

(i) When NaOH is added to the given soln., Zinc nitrate [Zn(NO₃)₂] reacts to form a gelatinous white ppt. which dissolves in excess of NaOH whereas, Ca(NO₃)₂ forms a milky white ppt. which is insoluble in excess of NaOH. Hence, the two can be distinguished.

(ii) Add silver nitrate soln. to the given solns., Magnesium chloride reacts to form a white ppt. which is soluble in NH₄OH and insoluble in dil. HNO₃. The other solution is Magnesium nitrate.

MgCl₂ + 2AgNO₃ ⟶ 2AgCl ↓ [white ppt.] + Mg(NO₃)₂

Mg(NO₃)₂ + AgNO₃ ⟶ no white ppt.

Question 5(2020)

Identify the salts P, Q, R from the following observations:

(i) Salt P has light bluish green colour. On heating, it produces a black coloured residue. Salt P produces brisk effervescence with dil. HCl and the gas evolved turns lime water milky, but no action with acidified potassium dichromate solution.

(ii) Salt Q is white in colour. On strong heating, it produces buff yellow residue and liberates reddish brown gas. Solution of salt Q produces chalky white insoluble ppt. with excess of ammonium hydroxide.

(iii) Salt R is black in colour. On reacting with conc. HCl, it liberates a pungent greenish yellow gas which turns moist starch iodide paper blue black.

Answer

(i) Copper carbonate

(ii) Lead nitrate

(iii) Manganese dioxide

Additional Questions

Question 1

The following materials are provided – solutions of cobalt chloride, ammonia, potassium permanganate, lime water, starch-iodide, sodium hydroxide, lead acetate, potassium iodide. Also provided are litmus and filter papers, glowing splinters and glass rods. Using the above how would you distinguish between :

(a) a neutral, acidic and a basic gas

(b) oxygen and hydrogen gas

(c) carbon dioxide and sulphur dioxide gas

(d) chlorine and hydrogen chloride gas

(e) hydrogen sulphide and nitrogen dioxide gas

(f) ammonia and carbon dioxide gas

(g) zinc carbonate and potassium nitrate

(h) hydrated copper sulphate and anhydrous copper sulphate

(i) ammonium sulphate and sodium sulphate.

Answer

(a) Litmus paper test is done to distinguish between neutral, acidic and basic gas. Neutral gas does not effect litmus paper. Acidic gas will turn blue litmus paper red and basic gas will turn red litmus blue.

(b) A burning wooden splinter is extinguished in hydrogen whereas oxygen gas rekindles a glowing wooden splinter. Hence, the two gases can be distinguished using a wooden splinter.

(c) There is no effect of carbon dioxide (CO₂) gas on potassium permanganate soln. whereas sulphur dioxide (SO₂) turns acidified potassium permanganate from pink to clear colourless.

(d) Chlorine turns moist blue litmus red and then bleaches it.

Cl₂ + H₂O ⟶ HOCl + HCl

HOCl ⟶ HCl + [O]

Colouring matter + [O] ⟶ Colourless or bleached product.

Whereas, HCl gas only turns moist blue litmus paper red and does not bleach it.

(e) Nitrogen dioxide (NO₂) liberates iodine [violet vapours] with potassium iodide KI soln., and turns potassium iodide paper brown.

2KI + 2NO₂ ⟶ 2KNO₂ + I₂

Hydrogen sulphide (H₂S) gas turns potassium permanganate (KMnO₄) from pink to colourless.

2KMnO₄ + 3H₂SO₄ + 5H₂S ⟶ K₂SO₄ + 2MnSO₄ + 8H₂O + 5S

(f) In the presence of ammonia moist red litmus turns blue and carbon dioxide turns moist blue litmus faint red.

(g) If we heat the given salts, zinc carbonate turns yellow, evolving carbon dioxide gas which turns lime water milky and has no effect on potassium permanganate solution.

$\begin{matrix} \text{ZnCO}_3 &\xrightarrow\Delta & \text{ZnO} & + & \text{CO}_2 \\ \text{[Zinc} & & \text{[Zinc oxide]} & & \text{[Carbon} \\ \text{carbonate]} & & \text{[yellow - hot]} & & \text{dioxide]} \\ \text{[white]} & & \text{[white - cold]} \end{matrix}$

Whereas, potassium nitrate emits oxygen gas on heating. Oxygen gas rekindles a glowing wooden splinter.

2KNO₃ ⟶ 2KNO₂ + O₂

Hence, the two salts can be distinguished.

(h) Hydrated copper sulphate is bright blue in colour, whereas anhydrous copper sulphate appears as a white powder. Hence, the two can be distinguished by their colour and appearance.

(i) When ammonium sulphate is heated with NaOH, ammonia gas will be produced which turns red litmus blue whereas, sodium sulphate will not react with sodium hydroxide.

(NH₄)₂SO₄ + 2NaOH ⟶ Na₂SO₄ + 2H₂O + 2NH₃

Na₂SO₄ + NaOH ⟶ no reaction.

Question 2

Give a chemical test to distinguish between the following:

(i) Sodium carbonate and sodium sulphate

(ii) Potassium chloride and potassium nitrate

(iii) Copper carbonate and copper sulphite

(iv) Lead chloride and lead sulphide

(v) Iron (II) sulphate and iron (III) sulphate

(vi) Calcium sulphate and zinc sulphate

(vii) Lead nitrate and zinc nitrate

(viii) Copper sulphate and calcium sulphate

(ix) Manganese dioxide and copper (II) oxide

(x) dil. HCl, dil. HNO₃, dil. H₂SO₄.

[explain the procedure for the preparation of the solutions for the above tests wherever required]

Answer

(i) When BaCl₂ solution is added to sodium carbonate, a white ppt. is formed which is soluble in dil. HCl.

Na₂CO₃ + BaCl₂ ⟶ BaCO₃ ↓ [white ppt.] + 2NaCl

BaCO₃ + 2HCl ⟶ BaCl₂ [soluble] + H₂O + CO₂

When BaCl₂ solution is added to sodium sulphate, a white ppt. is formed which is insoluble in dil. HCl.

Na₂SO₄ + BaCl₂ ⟶ BaSO₄ ↓ [white ppt.] + 2NaCl

Hence, the two compound can be distinguished.

(ii) Add silver nitrate soln. to the given solns., potassium chloride reacts to form a white ppt. which is soluble in NH₄OH and insoluble in dil. HNO₃. The other soln. is potassium nitrate.

KCl + AgNO₃ ⟶ AgCl + KNO₃

KNO₃ + AgNO₃ ⟶ no white ppt.

(iii) Add dil. H₂SO₄ to both the solns. and heat. When a colourless, odourless gas is evolved which has no effect on acidified KMnO₄ or K₂Cr₂O₇ solns., then the gas is carbon dioxide and the compound is copper carbonate.

When a colourless gas with a suffocating odour is evolved which turns pink acidified KMnO₄ to colourless then the gas is sulphur dioxide and the compound is copper sulphite.

(iv) Add dil. H₂SO₄ to both the solns. and heat. When a colourless gas with a smell of rotten eggs is evolved which turns pink acidified KMnO₄ colourless, then the gas is hydrogen sulphide and the compound is lead sulphide.

Now, add conc. H₂SO₄ to both the solns. and heat. When a gas with a pungent smell is evolved, which gives dense white fumes when a rod dipped in ammonia soln. is brought near it then the gas is HCl and compound is lead chloride.

(v) When sodium hydroxide is added to the two solns., Iron (II) sulphate solution gives a dirty green ppt. of Fe(OH)₂ whereas, Iron (III) sulphate solution forms a reddish brown ppt. of Fe(OH)₃. Hence, the two compounds can be distinguished.

(vi) When NaOH is added to the given soln., zinc sulphate reacts to form a gelatinous white ppt. which dissolves in excess of NaOH.

ZnSO₄ + 2NaOH ⟶ Na₂SO₄ + Zn(OH)₂ ↓

Zn(OH)₂ + 2NaOH [excess] ⟶ 2H₂O + Na₂ZnO₂

Whereas, calcium sulphate forms a milky white ppt. which is insoluble in excess of NaOH. Hence, the two can be distinguished.

CaSO₄ + 2NaOH ⟶ Ca(OH)₂ ↓ + Na₂SO₄

(vii) When NaOH is added to each of the compounds, lead nitrate forms a chalky white precipitate of lead hydroxide [Pb(OH)₂]

Pb(NO₃)₂ + 2NaOH ⟶ 2NaNO₃ + Pb(OH)₂ ↓

Whereas a gelatinous white precipitate of zinc hydroxide [Zn(OH)₂] is formed in case of zinc nitrate.

Zn(NO₃)₂ + 2NaOH ⟶ 2NaNO₃ + Zn(OH)₂ ↓

(viii) When NaOH is added to each of the compounds, copper sulphate forms a pale blue precipitate of copper [II] hydroxide [Cu(OH)₂]

CuSO₄ + 2NaOH ⟶ Na₂SO₄ + Cu(OH)₂ ↓

Whereas a milky white precipitate of calcium hydroxide [Ca(OH)₂] is formed in case of calcium sulphate.

CaSO₄ + 2NaOH ⟶ Na₂SO₄ + Ca(OH)₂ ↓

(ix) When each of the compound is heated with conc. hydrochloric acid, greenish yellow (chlorine) gas is evolved in case of manganese dioxide and filtrate is brownish in colour. Whereas, no chlorine gas is evolved in case of copper (II) oxide and filtrate is bluish in colour.

MnO₂ + 4HCl ⟶ MnCl₂ + 2H₂O + Cl₂

CuO + 2HCl ⟶ CuCl₂ + H₂O

(x) When BaCl₂ is added to the three acids, dil sulphuric acid reacts with BaCl₂ to give a white ppt. of BaSO₄ but with dil HCl and dil HNO₃ no white ppt. is produced.

BaCl₂ + H₂SO₄ ⟶ BaSO₄ ↓ [white ppt.] + 2HCl

To distinguish between dil HCl and dil HNO₃, we add Silver Nitrate (AgNO₃) solution to the two acids. Dil. HCl reacts with AgNO₃ to give a curdy white ppt. of Silver chloride (AgCl) but with dil HNO₃, no white ppt. is produced.

HCl + AgNO₃ ⟶ AgCl + HNO₃

HNO₃ + AgNO₃ ⟶ no white ppt.

Hence, the acids can be distinguished.

Question 3

Identify the cation [positive ion] and anion [negative ion] in - A, B and C. Also identify P, Q, R, S, T, U, V, W.

(a) Substance 'A' is water soluble and gives a curdy white precipitate 'P' with silver nitrate solution. 'P' is soluble in ammonium hydroxide but insoluble in dil. HNO₃. Substance 'A' reacts with ammonium hydroxide solution to give a white precipitate 'Q' soluble in excess of NH₄OH.

(b) A solution of substance 'B' is added to barium chloride solution. A white ppt. 'R' is formed, insoluble in dil. HCl or HNO₃. A dirty green ppt. 'S' is formed on addition of ammonium hydroxide to a solution of 'B' and the precipitate is insoluble in excess of ammonium hydroxide.

(c) Substance 'C' is a coloured, crystalline salt which on heating decomposes leaving a black residue 'T'. On addition of copper turnings and conc. H₂SO₄ to 'C' a coloured acidic gas 'U' is evolved on heating. A solution of 'C' is added to NaOH soln. until in excess. A pale blue ppt. 'P' is obtained insoluble in excess of NaOH. A solution of 'C' is then added to NH₄OH soln. in excess to give an inky blue solution 'V'. A solution of 'C' is warmed and hydrogen sulphide gas is passed through it. A black ppt. 'W' appears.

Answer

(a) A is ZnCl₂

Cation and anion in ZnCl₂ : Zn²⁺ and Cl^-

ZnCl₂ + 2AgNO₃ ⟶ 2AgCl ↓ [curdy white ppt.] + Zn(NO₃)₂

P is AgCl

The precipitate AgCl is soluble in ammonium hydroxide but insoluble in dil. HNO₃.

ZnCl₂ + 2NH₄OH ⟶ 2NH₄Cl + Zn(OH)₂ ↓ [white ppt.]

2NH₄Cl + Zn(OH)₂ + 2NH₄OH [excess] ⟶ 4H₂O + [Zn(NH₃)₂]Cl₂

Q is Zn(OH)₂

(b) B is ferrous sulphate (FeSO₄)

Cation and anion in FeSO₄ : Fe²⁺ and SO₄^2-

FeSO₄ + BaCl₂ ⟶ BaSO₄ ↓ [white ppt.] + 2NaCl

R is BaSO₄

FeSO₄ + 2NH₄OH ⟶ (NH₄)₂SO₄ + Fe(OH)₂ ↓

S is Fe(OH)₂

(c) C is copper nitrate Cu(NO₃)₂

Cation and anion in Cu(NO₃)₂ : Cu²⁺ and NO₃^1-

Black residue (T) : CuO - copper oxide

$\underset{\text{[Copper nitrate (blue)]}}{2\text{Cu(NO}_3)_2} \overset{\Delta}{\longrightarrow} \underset{\text{[copper oxide - black]}}{2\text{CuO}} + \underset{\text{[Nitrogen dioxide]}}{4\text{NO}_2} + \underset{\text{[Oxygen]}}{\text{O}_2}$

U is Nitrogen dioxide (NO₂)

P is Copper [II] hydroxide Cu(OH)₂

Cu(NO₃)₂ + 2NaOH ⟶ 2NaNO₃ + Cu(OH)₂ ↓

V is [Cu(NH₃)₄]SO₄

Cu(NO₃)₂ + 2NH₄OH ⟶ Cu(OH)₂ ↓ + 2NH₄NO₃

2Cu(OH)₂ + 2NH₄NO₃ + 2NH₄OH ⟶ 2Cu(NH₃)₄NO₃ + 4H₂O

W is CuS

Cu(NO₃)₂ + H₂S ⟶ CuS + 2HNO₃

Unit Test Paper — Chemistry Practicals

Question 1

Match the 'cations' A to F and the solubility of ppt. G or H with the correct colours from 'X' and 'Y'.

Answer

'X' on addition of NaOH:

'Y' on addition of NH₄OH:

Question 2

Select the correct 'anion' of a salt from the anions given, which matches with description 1 to 5.

A: CO₃^2-, B: NO₃^1-, C: SO₄^2-, D: Cl^-, E: S^2-

1. The salt soln. reacts with AgNO₃ soln. to give a white ppt. insoluble in dil. HNO₃.
2. The salt soln. reacts with Ba(NO₃)₂ soln. to give a white ppt. insoluble in dil. HNO₃.
3. The salt soln. reacts with Ba(NO₃)₂ soln. to give a white ppt. soluble in dil. HNO₃ but insoluble in dil. H₂SO₄.
4. The salt reacts with dil. H₂SO₄ on heating evolving a gas which turns KMnO₄ soln. pink to colourless.
5. The salt reacts with conc. H₂SO₄ on heating evolving a coloured gas which turns potassium iodide paper brown.

Answer

1. D: Cl^-
2. C: SO₄^2-
3. A: CO₃^2-
4. E: S^2-
5. B: NO₃^1-

Question 3

Give balanced equations for the conversions A and B.

1. $\text{Metallic carbonate} \quad \raisebox{0.35em}{A} \space \xrightarrow{\text{Ba(NO}_3)_2} \text{White precipitate} \quad \raisebox{0.35em}{B} \space \xrightarrow{\text{dil. HCl}} \text{precipitate dissolves}$
2. $\text{Metallic sulphide} \quad \raisebox{0.35em}{A} \space \xrightarrow{\text{Pb(CH}_3{\text{COO}})_2 \text{ soln.}} \text{black precipitate }$
3. $\text{Metallic salt} \quad \raisebox{0.35em}{A} \space \xrightarrow{\text{BaCl}_2 \text{ soln.}} \text{Barium sulphite} \quad \raisebox{0.35em}{B} \space \xrightarrow{\text{dil. HCl}} \text{Barium chloride}$
4. $\text{Metallic chloride} \quad \raisebox{0.35em}{A} \space \xrightarrow{\text{conc. H}_2\text{SO}_4\Delta} \text{Gas evolved} \quad \raisebox{0.35em}{B} \space \xrightarrow{\text{AgNO}_3} \text{White precipitate}$
5. $\text{Metallic salt} \quad \raisebox{0.35em}{A} \space \xrightarrow{\text{BaCl}_2} \text{White precipitate insoluble in dil. HCl}$

Answer

1. Na₂CO₃ + Ba(NO₃)₂ ⟶ BaCO₃ ↓ [white ppt.] + 2NaNO₃
   BaCO₃ + 2HCl ⟶ BaCl₂ [soluble] + H₂O + CO₂

2. Pb(CH₃COO)₂ [colourless] + H₂S ⟶ PbS ↓ [black] + 2CH₃COOH

3. Na₂SO₃ + BaCl₂ ⟶ BaSO₃ ↓ [white ppt.] + 2NaCl
   BaSO₃ + 2HCl ⟶ BaCl₂ [soluble] + H₂O + SO₂

4. NaCl + H₂SO₄ [conc.] $\xrightarrow{\lt 200 \degree\text{C}}$ NaHSO₄ + HCl
   HCl + AgNO₃ ⟶ AgCl ↓ [white ppt.] + HNO₃

5. Na₂SO₄ + BaCl₂ ⟶ BaSO₄ ↓ [white ppt.] + 2NaCl

Question 4

Complete the table given below :

Answer

Question 5

Select the correct word from the words in bracket.

1. The solution which on heating with CaCO₃ evolves CO₂ gas. [conc. H₂SO₄ / dil.H₂SO₄ / dil. HCl]
2. The solution which can be used to distinguish an ammonium salt from a sodium salt. [CuCl₂ soln. / NH₄OH / dil. H₂SO₄ / AgNO₃ soln.]
3. The pH of blood is around 7.4, of saliva is 6.5 and of acid rain is around 4.5. The solution which is slightly alkaline of the three. [saliva / acid rain / blood]
4. Decomposition of [NaCl / NaHCO₃ / NaNO₃] by dil. H₂SO₄, forms an unstable acid.
5. A metal which reacts with an alkali to liberate hydrogen. [iron / copper / aluminium]

Answer

1. dil. H₂SO₄
2. CuCl₂ soln
3. blood
4. NaHCO₃
5. aluminium.