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Solutions for Chemistry, Class 10, ICSE
The three ways to identify ammonia gas are:
Give a chemical test to distinguish between :
(a) Sodium chloride soln. and sodium nitrate soln.
(b) Hydrogen chloride gas and hydrogen sulphide gas.
(c) Calcium nitrate soln. and zinc nitrate soln.
(d) Carbon dioxide gas and sulphur dioxide gas.
(e) Sodium carbonate and sodium sulphite
(f) Ferrous nitrate and lead nitrate.
(g) Manganese dioxide and copper (Il) oxide.
(h) Sodium nitrate and sodium sulphite [using dilute sulphuric acid].
(i) Sodium chloride and sodium sulphide
(j) Sodium hydroxide solution and ammonium hydroxide solution.
(k) Ammonium sulphate and sodium sulphate.
(l) Sulphuric acid from nitric acid and hydrochloric acid.
(m) Magnesium chloride and magnesium nitrate solution
(n) Zinc sulphate solution and zinc chloride solution
(o) Iron (II) chloride soln. and iron (III) chloride soln.
(p) Calcium nitrate soln. and calcium chloride soln.
(a) 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.
(b) 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
(c) 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.
(d) 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
(e) 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.
(f) 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.
(g) 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
(h) 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.
(i) When dil. sulphuric acid is added to sodium sulphide and heated, hydrogen sulphide gas is evolved which has the smell of rotten eggs and turns lead acetate paper black.
Whereas, when conc. sulphuric acid is added to sodium chloride and heated, HCl gas is evolved which gives dense white fumes when a glass rod dipped in ammonia is brought near it.
(j) When ammonium hydroxide is reacted with a solution of copper sulphate, a pale blue ppt. of copper hydroxide is formed which dissolves in excess of ammonium hydroxide forming a deep blue solution of a soluble complex salt [Cu(NH3)4]SO4 [tetramine copper [II] sulphate] .
CuSO4 + 2NH4OH ⟶ (NH4)2SO4 + Cu(OH)2 ↓
Cu(OH)2 + (NH4)2SO4 + 2NH4OH ⟶ [Cu(NH3)4]SO4 + 4H2O
Whereas, when copper sulphate is added to sodium hydroxide solution, pale blue ppt. of copper hydroxide is obtained which is insoluble in excess of sodium hydroxide.
CuSO4 + 2NaOH ⟶ Na2SO4 + Cu(OH)2 ↓
(k) When ammonium sulphate is heated with NaOH, ammonia gas is produced which turns red litmus blue whereas, sodium sulphate does not react with sodium hydroxide.
(NH4)2SO4 + 2NaOH ⟶ Na2SO4 + 2H2O + 2NH3
Na2SO4 + NaOH ⟶ no reaction.
(l) Barium chloride reacts with dil H2SO4 to form a white ppt. of barium sulphate.
BaCl2 + H2SO4 [dil.] ⟶ 2HCl + BaSO4 ↓ [white ppt.]
Whereas, HCl and nitric acid does not form a white ppt.
BaCl2 + HCl ⟶ No white ppt.
BaCl2 + HNO3 ⟶ No white ppt
(m) Add silver nitrate solution to the given solns., magnesium chloride reacts to form a white ppt of silver chloride (AgCl). Whereas, magnesium nitrate show no reaction with silver nitrate. Hence, they can be distinguished using silver nitrate.
MgCl2 + 2AgNO3 ⟶ 2AgCl↓ + Mg(NO3)2
(n) 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.
(o) 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.
(p) 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.
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
Iron (II) sulphate
Reason
So, the salt must be an iron (II) salt (due to the pale green ppt) containing sulphate (due to white ppt with BaCl2), i.e., Iron (II) sulphate.
Ammonium sulphate
Reason — Ammonium sulphate reacts with sodium hydroxide to produce sodium sulphate, ammonia gas, and water. The red litmus turns blue due to the basic ammonia gas.
sodium hydroxide solution
Reason — Let's analyze each option:
Aqueous potassium chloride: This will precipitate lead(II) chloride (PbCl₂) from lead (II) nitrate, whereas zinc chloride is soluble. Thus, this can distinguish between them, as lead will form a precipitate while zinc will not.
Aqueous sodium sulphate: This will precipitate lead (II) sulphate (PbSO₄) from lead (II) nitrate since PbSO₄ is poorly soluble, while zinc sulphate is soluble. Thus, this can also distinguish between them.
Dilute sulphuric acid: Similar to aqueous sodium sulphate, adding dilute H₂SO₄ will precipitate lead (II) sulphate from lead (II) nitrate but not zinc sulphate from zinc nitrate. So, this can distinguish between them.
Sodium hydroxide solution: When added to both solutions in excess, this will not distinguish them because both lead (II) hydroxide and zinc hydroxide will initially precipitate but then both dissolve upon further addition of NaOH. Lead hydroxide forms a plumbate ion, and zinc hydroxide forms a zincate ion, both of which are soluble.
A student takes Cu, Al, Fe and Zn strips, separately in four test tubes labeled as I, II, III and IV respectively. He adds 10 ml of freshly prepared ferrous sulphate solution to each test tube and observes the colour of the metal residue in each case.
He would observe a black residue in the test tubes:
(II) and (IV)
Reason — Zinc and aluminium are more reactive than iron. Therefore they will displace iron which will be seen as black residue.
copper nitrate
Reason — When copper nitrate reacts with ammonium hydroxide NH4OH, it initially forms a light blue precipitate of copper(II) hydroxide (Cu(OH)2).
This precipitate is soluble in excess ammonium hydroxide due to the formation of a deep blue complex, tetraamminecopper(II) ([Cu(NH3)4]2+), making it soluble.
Dilute sulphuric acid is added to sodium sulphite. The following observations are noticed:
P — A gas with a suffocating smell is evolved which turns lime water milky.
Q — A foul smelling gas is evolved which turns lead acetate solution black.
R — A gas is evolved which decolourises pink KMnO4
Which of the following is true?
Both P and R
Reason — When dilute sulphuric acid is added to sodium sulphite, a colourless gas with a smell of burning sulphur i.e., suffocating odour is evolved.
Na2SO3 + H2SO4 ⟶ Na2SO4 + H2O + SO2
This gas turns lime water milky.
Ca(OH)2 + SO2 ⟶ CaSO3↓ + H2O
This gas also decolourises pink KMnO4.
2KMnO4 + 2H2O + 5SO2 ⟶ K2SO4 + 2MnSO4 + 2H2SO4
Both P and R
Reason — Molecular compounds usually have weak intermolecular forces, leading to low boiling points. Molecular compounds typically do not form ions in solution (unless they’re acids or bases). So their solutions are usually non-conducting. Most covalent compounds are insoluble in water, unless they are polar.
Assertion (A): HCl can be tested by AgNO3 solution.
Reason (R): White precipitate of AgCl is formed when AgNO3 is added to HCl.
Both A and R are true and R is the correct explanation of A.
Explanation— HCl can be tested by AgNO3 solution. When, HCl gas is passed through Silver nitrate solution white precipitate of AgCl is formed.
AgNO3(aq) + HCl ⟶ AgCl↓ + HNO3
Hence, both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
Assertion (A): SO2 and CO2 can be distinguished by lime water.
Reason (R): Both turn lime water milky.
A is false but R is true.
Explanation— SO2 and CO2 cannot be distinguished by lime water, because both turn lime water milky. Hence, the assertion (A) is false.
Ca(OH)2 + SO2 ⟶ CaSO3↓ + H2O
Ca(OH)2 + CO2 ⟶ CaCO3↓ + H2O
SO2 and CO2 turn lime water milky. Hence, the reason (R) is true.
Assertion (A): On passing H2S through lead nitrate solution, a black precipitate is formed.
Reason (R): Lead nitrate reacts with H2S to form a black precipitate of PbS.
Both A and R are true and R is the correct explanation of A.
Explanation— H2S is colourless gas having a foul smell of rotten eggs. When H2S is passed through lead nitrate solution, a black precipitate of PbS is formed.
Pb(NO3)2 + H2S ⟶ PbS ↓ + 2HNO3
Hence, both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
Assertion (A): Chloride can be tested by adding concentrated sulphuric acid to it. A colourless gas with a pungent odour is evolved.
Reason (R): The gas formed is chlorine.
A is true but R is false.
Explanation— The presence of chloride ion can be tested by adding concentrated sulphuric acid. It is observed that the colourless HCl gas of pungent odour is evolved. Hence, the assertion (A) is true.
The colourless gas evolved in the above reaction is HCl. Hence, the reason (R) is false.
Assertion (A): If barium nitrate solution is added to sodium sulphate solution, a white precipitate is formed.
Reason (R): Sodium sulphate solution is white in colour.
A is true but R is false.
Explanation— When barium nitrate solution is added to sodium sulphate salt solution, white precipitate which is insoluble in water is formed. White precipitate is soluble in dil. HCl. Hence, the assertion (A) is true.
Sodium sulphate solution is a colourless solution. Hence, the reason (R) is false.
Name the anion present in each of the following compounds.
(a) Compound A when warmed with concentrated sulphuric acid gives a gas which fumes in moist air and which gives dense white fumes with ammonia.
(b) When barium chloride solution is added to a solution of compound B, a white precipitate insoluble in dilute hydrochloric acid is formed.
(c) The action of heat on the insoluble compound C produces a gas which turns lime water turbid.
(d) Compound D when warmed with dilute sulphuric acid gives a gas which turns acidified dichromate solution green.
(a) Cl-
(b) SO42-
(c) CO32-
(d) SO32-
A given white crystalline salt was tested as follows :
(a) It dissolved in water and the resulting solution of the salt turned blue litmus red.
(b) Addition of barium chloride solution into this solution gave a white precipitate.
(c) A flame test on the salt gave a persistent golden-yellow colourisation.
What conclusion can be drawn for each observation?
(a) As the salt solution turned blue litmus red hence the salt may be an acid.
(b) As white ppt. is obtained on addition of barium chloride so the salt may contain SO42-, SO32-, CO32-.
(c) Persistent golden yellow colourisation which suggests presence of Na+ ion.
(a) Sodium hydroxide solution is added to solution A. A white precipitate is formed which is insoluble in excess sodium hydroxide solution. Name the metal ion present in solution A.
(b) When ammonium hydroxide is added to solution B, a pale blue precipitate is formed. This pale blue precipitate dissolves in excess ammonium hydroxide giving an inky blue solution. Name the cation present in solution B.
(a) Ca2+
(b) Cu+
The questions (i) to (v) refer to the following salt solutions listed A to F.
A. Copper nitrate
B. Iron (II) sulphate
C. Iron (III) chloride
D. Lead nitrate
E. Magnesium sulphate
F. Zinc chloride
(i) Which two solutions will give a white precipitate when treated with dilute hydrochloric acid followed by barium chloride solution?
(ii) Which two solutions will give a white precipitate when treated with dilute nitric acid followed by silver nitrate solution?
(iii) Which solution will give a white precipitate, when either dilute hydrochloric acid or dilute sulphuric acid is added to it?
(iv) Which solution becomes a deep/inky blue colour when excess of ammonium hydroxide is added to it?
(v) Which solution gives a white precipitate with excess ammonium hydroxide solution?
(i) Iron (II) Sulphate and Magnesium sulphate
(ii) Iron (III) chloride and Zinc Chloride
(iii) Lead nitrate
(iv) Copper nitrate.
(v) Lead nitrate.
Salts A, B, C, D and E undergo reactions (i) to (v) respectively. Identify the anion present in these salts on the basis of these reactions.
(a) When silver nitrate solution is added to a solution of A, a white precipitate, insoluble in dilute nitric acid, is formed.
(b) Addition of dilute hydrochloric acid to B produces a gas which turns lead acetate paper black.
(c) When a freshly prepared solution of ferrous sulphate is added to a solution of C and concentrated sulphuric acid is gently poured from the side of the test-tube, a brown ring is formed.
(d) When dilute sulphuric acid is added to D, a gas is produced which turns acidified potassium dichromate solution from orange to green.
(e) Addition of dilute hydrochloric acid to E produces an effervescence. The gas produced turns lime water milky but does not affect acidified potassium dichromate solution.
(a) Chloride ion (Cl-)
(b) Sulphide ion (S2-)
(c) Nitrate (NO3-)
(d) Sulphite ion (SO32-)
(e) Carbonate ion (CO32-)
State your observations when:
(a) lead nitrate solution and sodium chloride solution are mixed.
(b) zinc chloride solution, zinc nitrate solution and zinc sulphate solutions are added individually to
(i) barium chloride solution,
(ii) lead nitrate solution.
(c) Decomposition of bicarbonates by dil. H2SO4
2NaHCO3 + H2SO4 ⟶ Na2SO4 + 2H2O + 2CO2
2KHCO3 + H2SO4 ⟶ K2SO4 + 2H2O + 2CO2
(d) Zinc carbonate is heated strongly.
(e) a piece of moist blue litmus paper is introduced in a jar of chlorine.
(f) paper soaked in potassium permanganate solution — is introduced in each case into a jar of sulphur dioxide.
(g) When moist starch iodide paper is introduced into chlorine gas.
(a) When lead nitrate solution is mixed with sodium chloride solution, a white precipitate of lead chloride is seen and the soluble salt sodium nitrate is formed.
2NaCl + Pb(NO3)2 ⟶ 2Na(NO3)2 + PbCl2↓
Precipitate of lead chloride and sodium nitrate are formed.
(b)
| Zinc Chloride Solution | Zinc Nitrate Solution | Zinc Sulphate Solution | |
|---|---|---|---|
| Barium Chloride Solution | No reaction | No ppt. observed | White ppt. is obtained which is insoluble in mineral acid. |
| Lead Nitrate Solution | White ppt. is obtained | No reaction | No ppt. observed |
(c) Carbon dioxide gas is evolved with brisk effervescence which turns blue litmus red and extinguishes a burning wooden splinter.
(d) When zinc carbonate is strongly heated, the light amorphous white solid, changes to pale yellow colour.
ZnCO3 ZnO + CO2
(e) Moist blue litmus turns red and then may eventually get bleached.
(f) Sulphur dioxide turns potassium permanganate from pink to clear colourless.
2KMnO4 + 2H2O + 5SO2 ⟶ K2SO4 + 2MnSO4 + 2H2SO4
(g) Chlorine gas turns moist starch iodide paper blue black.
Cl2 + 2KI ⟶ 2KCl + I2
[Starch + I2 ⟶ blue black colour]
| Hydrogen sulphide | Ammonia | Sulphur dioxide | Hydrogen chloride | |
|---|---|---|---|---|
| Shake the gas with red litmus solution | No change | Red litmus becomes blue | No change | No change |
| Shake the gas with blue litmus solution | Blue litmus becomes red | No change | Blue litmus becomes red | Blue litmus becomes red |
| Apply a burning splint to the gas | Burning splint is extinguished | Burning splint is extinguished | Burning splint is extinguished | Burning splint is extinguished |
Identify the anion present in the following compounds :
(a) Compound X on heating with copper turnings and conc. sulphuric acid liberates a reddish brown gas.
(b) 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.
(c) 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.
(d) Compound L on reacting with barium chloride solution gives a white precipitate insoluble in dilute hydrochloric acid or dilute nitric acid.
(a) Nitrate ion, NO3-
(b) Chloride ion, Cl-
(c) Carbonate ion, CO32-
(d) Sulphate ion, SO42-
(a) Zinc and aluminium cannot be distinguished by heating the metal powder with concentrated sodium hydroxide solution as they both react with conc. alkalis to form soluble sodium salts and hydrogen gas.
Zinc reacts to form sodium zincate
Zn + 2NaOH ⟶ H2 + Na2ZnO2 [sodium zincate]
Aluminium reacts to form sodium aluminate
2Al + 2NaOH + 2H2O ⟶ 3H2 + 2NaAlO2 [sodium aluminate]
(b) Yes, calcium nitrate and lead nitrate can be distinguished using ammonium hydroxide solution. Ammonium hydroxide on reaction with lead nitrate gives chalky white precipitate of Pb(OH)2. No precipitation occurs on adding Ammonium hydroxide to calcium nitrate even when it is added in excess.
Pb(NO3)2 + 2NH4OH ⟶ Pb(OH)2 + 2NH4NO3
The substance 'R' is Ferrous sulphate (FeSO4.7H2O). 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 FeSO4 + 7H2O
2FeSO4 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.
Match the gases in column I to the identification of the gases mentioned in column II
| Column I | Column II |
|---|---|
| (a) Hydrogen sulphide | A. Turns acidified potassium dichromate solution green |
| (b) Nitric oxide | B. turns lime water milky |
| (c) Carbon dioxide | C. Turns reddish brown when it reacts with oxygen |
| (d) Sulphur dioxide | D. Turns moist lead aceate paper silvery black |
| Column I | Column II |
|---|---|
| (a) Hydrogen sulphide | D. Turns moist lead aceate paper silvery black |
| (b) Nitric oxide | C. Turns reddish brown when it reacts with oxygen |
| (c) Carbon dioxide | B. Turns lime water milky |
| (d) Sulphur dioxide | A. Turns acidified potassium dichromate solution green |
(a) Ammonia
(b) Hydrogen sulphide
(c) Sulphur dioxide
From the following list of substances choose those which meet the description given below.
Two compounds whose aqueous solutions give white precipitates with dilute hydrochloric acid.
Silver nitrate and Lead nitrate.
HCl + AgNO3 ⟶ AgCl + HNO3
2HCl + Pb(NO3)2 ⟶ PbCl2 + 2HNO3
Use the information given in (a) to (h) to identify the substances P to W selecting your answers from the given list.
| List | ||
|---|---|---|
| Calcium | Oxygen | Copper (II) oxide |
| Carbon | Calcium hydroxide | Copper (II) nitrate |
| Lead (II) oxide | Hydrogen chloride | Chlorine |
| Lead (II) nitrate | Calcium oxide | Ammonium chloride |
(a) P is a white solid. When heated produces white fumes (sublime).
(b) P and R on warming produce an alkaline gas.
(c) On adding water to T, heat is evolved and R is formed.
(d) Q burns brightly in air to form T.
(e) When S is heated, it gives off brown fumes and leaves a black residue of U.
(f) A solution of S is formed by warming U with dilute nitric acid.
(g) V is a gaseous non-metallic element that reacts with hydrogen to form W.
(h) A solution of W will neutralise the solution of R.
(P) Ammonium chloride
(Q) Calcium
(R) Calcium hydroxide
(S) Lead (II) Nitrate
(T) Calcium Oxide
(U) Lead (II) Oxide
(V) Chlorine
(W) Hydrogen chloride
| Carbonate | Colour of residue on cooling |
|---|---|
| Zinc carbonate | White |
| Lead carbonate | Yellow |
| Copper carbonate | Black |
Sodium hydroxide solution is added first in a small quantity, then in excess to the aqueous salt solutions of copper (II) sulphate, zinc nitrate, lead nitrate, calcium chloride and iron (III) sulphate. Copy the following table and write the colour of the precipitate in (i) to (v) and the nature of the precipitate (soluble or insoluble) in (vi) to (x).
| Aqueous salt solution | Colour of precipitate when NaOH is added in a small quantity | Nature of precipitate (soluble or insoluble) when NaOH is added in excess |
|---|---|---|
| Copper (II) sulphate | (i) | (vi) |
| Zinc nitrate | (ii) | (vii) |
| Lead nitrate | (iii) | (viii) |
| Calcium chloride | (iv) | (ix) |
| Iron (III) sulphate | (v) | (x) |
| Aqueous salt solution | Colour of precipitate when NaOH is added in a small quantity | Nature of precipitate (soluble or insoluble) when NaOH is added in excess |
|---|---|---|
| Copper (II) sulphate | (i) Pale blue | (vi) Insoluble |
| Zinc nitrate | (ii) White gelatinous | (vii) Soluble |
| Lead nitrate | (iii) White chalky | (viii) Soluble |
| Calcium chloride | (iv) White curdy | (ix) Insoluble |
| Iron (III) sulphate | (v) Reddish brown | (x) Insoluble |
| Solution | Acids | Alkalies |
|---|---|---|
| (a) Alkaline phenolphthalein solution, | Colourless | Pink |
| (b) Methyl orange solution | Pink | Yellow |
| (c) Neutral litmus solution | Red | Blue |
(a) Select the correct answer from A, B, C and D.
A. Nitroso iron (II) sulphate
B. Iron (III) chloride
C. Chromium sulphate
D. Lead (II) chloride.
(a) The compound which is responsible for the green colour formed when SO2 is bubbled through acidified potassium dichromate solution.
(b) Compound responsible for brown ring.
(a) Chromium sulphate
K2Cr2O7 + H2SO4 + 3SO2 ⟶ K2SO4 + Cr2(SO4)3 + H2O
(b) Nitroso iron (II) sulphate
A student was asked to perform two experiments in the laboratory based on the instructions given:
Observe the picture given below and state one observation for each of the Experiments 1 and 2 that you would notice on mixing the given solutions.
(a) Experiment 1
(b) Experiment 2
(a) White precipitate of barium sulphate is formed.
Reason — When a solution of zinc sulphate is added to a solution of barium chloride, a double displacement reaction occurs. This reaction involves the exchange of ions between the two compounds and a white precipitate of barium sulphate is formed.
ZnSO4 (aq) + BaCl2 (aq) ⟶ BaSO4 (s) + ZnCl2 (aq).
(b) Blue precipitate dissolves to form an inky blue/deep blue solution.
Reason — The light blue precipitate of copper(II) hydroxide (Cu(OH)2) dissolves in excess of ammonium hydroxide due to the formation of a deep blue complex, tetraamminecopper(II) ([Cu(NH3)4]2+), making it soluble.
(a) Give only one suitable chemical test to identify the following gases.
(i) Ammonia
(ii) Sulphur dioxide
(iii) Hydrogen chloride
(iv) Chlorine
(v) Carbon dioxide
(vi) Oxygen
(vii) Hydrogen
(b) Select a basic gas mentioned in Q. 1(a). How is the basic nature suspected?
(c) Select acidic gases from the gases mentioned in Q. 1 (a). How is the acidic nature suspected?
(d) The two gases A and B are bleaching agents. A is greenish yellow and bleaches due to it's oxidising property while B a colourless gas bleaches due to reduction. Identify A and B.
(e) Which gas turns blue cobalt chloride paper light pink?
Give one similarity in test between (i) Cl2 and HCl (ii) SO2 and CO2.
(i) Ammonia — When a glass rod dipped in conc. HCl is brought near the gas, dense white fumes of ammonium chloride are formed.
(ii) Sulphur dioxide — Sulphur dioxide gas turns acidified potassium permanganate from pink to clear colourless and acidified potassium dichromate from orange to clear green.
(iii) Hydrogen chloride — Forms a curdy white ppt. on passage through AgNO3 solution. The precipitate dissolves in excess of NH4OH.
(iv) Chlorine — Chlorine turns moist blue litmus red and then bleaches it.
(v) Oxygen — Oxygen gas rekindles a glowing wooden splinter.
(iv) Hydrogen — Hydrogen gas burns with a 'pop' sound in air.
(b) Ammonia is a basic gas. It turns red litmus blue.
(c) Sulphur dioxide, Hydrogen chloride, Chlorine, Carbon dioxide.
(d) A is Chlorine and B is Sulphur dioxide.
(e) Water vapour
(i) Similarity in test of Cl2 and HCl — Forms a white ppt. on reaction with AgNO3 solution.
(ii) Similarity in test of SO2 and CO2 — Both turn lime water milky.
(a) NH3, HCl, SO2, H2S, CO2, NO2, Cl2.
(b) Ammonia [NH3]
(c) Water vapour, hydrogen, oxygen
(d) sulphur dioxide [SO2]
(a) Sodium carbonate [Na2CO3] and potassium carbonate [K2CO3]
(b) Sulphur dioxide [SO2]
(c) Carbon dioxide [CO2]
(d) Chlorine [Cl2]
(e) Hydrogen sulphide [H2S]