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Class 10 Chemistry Chapter 18 Salts Test 8 | Punjab Board 2027
Chemistry Full chapter Test unit 18 Class 10
Prepare for your annual board examinations with the Class 10 Chemistry Chapter 18: Salts Class Test 8, part of the Punjab Board 2027 Model Papers and chapter-wise test series. This full unit test is designed to help students revise the complete Salts chapter according to the new syllabus issued by PECTAA Punjab. Students can assess their understanding of important chemical concepts, practise exam-style questions, identify areas for improvement, and strengthen their Chemistry preparation. Attempt this Class 10 Chemistry Chapter 18 test to improve your subject knowledge and prepare confidently for the Punjab Board Class 10 Chemistry annual examination 2027.
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Full Chapter Test
T-8
CHAPTER # 18
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Note:Four possible answers A, B, C and D to each question are given. The choice which you think is correct, fill that circle in front of that question with Marker or Pen ink. Cutting or filling two or more circles will result in zero mark in that question.
1.A salt is composed of:
(a) Only positive ions(b) Only negative ions(c) Positive and negative ions(d) Neutral atoms
2.Ionic compounds usually exist as:
(a) Liquids(b) Gases(c) Crystalline solids(d) Plasma
3.In NaCl, each Na+ ion is surrounded by:
(a) 4 Cl– ions(b) 6 Cl– ions(c) 8 Cl– ions(d) 2 Cl– ions
4.During melting of an ionic compound:
(a) Ions are destroyed(b) Electrostatic forces are broken(c) New ions are formed(d) Charges disappear
5.When ionic compounds dissolve in water, they:
(a) Form atoms(b) Dissociate into ions(c) Evaporate(d) Become neutral
6.Which of the following is an insoluble chloride?
(a) NaCl(b) KCl(c) AgCl(d) NH4Cl
7.Which salt can be prepared by titration method?
(a) AgCl(b) BaSO4(c) KNO3(d) CaCO3
8.Which of the following can act as an insoluble base?
(a) NaOH(b) KOH(c) CuO(d) NH4OH
9.Which base is not soluble in water?
(a) KOH(b) Mg(OH)2(c) NaOH(d) Na2CO3
10.The shape of the crystal of NaCl is:
(a) Cubic(b) Hexagonal(c) Rhombic(d) Trigonal
11.Which salts are always soluble in water?
(a) Chlorides(b) Sulphates(c) Nitrates(d) Carbonates
12.Water insoluble salt is:
(a) Calcium sulphate(b) Sodium sulphate(c) Potassium sulphate(d) Magnesium sulphate
Answers: 1-c 2-c 3-b 4-b 5-b 6-c 7-c 8-c 9-b 10-a 11-c 12-a
Here is Subjective Part of Paper.
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Marks: 48
CHEMISTRY (SUBJECTIVE TYPE)
Time: 1:45 minute
Part – I
2.Write short answers to any Five (5) questions:[5×2=10]
(i)Define a salt.
Ans: A salt is an ionic compound formed when the hydrogen of an acid is replaced by a metal (or ammonium) ion, e.g. NaCl, CuSO4.
(ii)What is a crystal lattice?
Ans: A crystal lattice is the regular, repeating three-dimensional arrangement of ions (or atoms/molecules) in a crystal.
(iii)Why do ionic compounds exist as crystalline solids?
Ans: Strong electrostatic forces between oppositely charged ions pull them into a fixed, orderly pattern, so they form hard crystalline solids.
(iv)Describe the arrangement of ions in NaCl.
Ans: Each Na+ ion is surrounded by six Cl− ions and each Cl− ion by six Na+ ions, forming a cubic lattice (6 : 6 coordination).
(v)Why common salt was used as commodity in the past?
Ans: Salt was scarce, valuable and essential for preserving food, so it was used in trade and as payment (the word “salary” comes from it).
(vi)What happens to ions when an ionic compound melts?
Ans: The lattice breaks down and the ions become free to move.
(vii)Why do ionic compounds not conduct electricity in solid state?
Ans: In the solid state the ions are held in fixed positions in the lattice and cannot move to carry charge.
(viii)How does an aqueous solution of an ionic compound conduct electricity?
Ans: In water the ions separate and become free and mobile; these moving ions carry electric charge through the solution.
3.Write short answers to any Five (5) questions:[5×2=10]
(i)Name three types of salts that are always soluble in water.
Ans: (1) All sodium salts, (2) all potassium salts, (3) all ammonium salts (also all nitrates).
(ii)Are all nitrates soluble? Explain.
Ans: Yes. All common nitrates (e.g. NaNO3, AgNO3, Pb(NO3)2) dissolve in water because water molecules overcome the forces between their ions.
(iii)Predict whether BaSO4 is soluble or insoluble. Give reason.
Ans: BaSO4 is insoluble. Most sulfates are soluble, but the sulfates of barium, lead and (sparingly) calcium are insoluble.
(iv)Write nuetralization reaction between sulfuric acid and sodium hydroxide.
Ans: H2SO4 + 2NaOH → Na2SO4 + 2H2O
(v)Name three types of insoluble bases used in salt formation.
Ans: Copper(II) oxide (CuO), zinc oxide (ZnO) and magnesium oxide (MgO).
(vi)Why must the base or metal be added in excess?
Ans: To make sure all the acid is used up. The unreacted excess solid is insoluble and is easily removed by filtration, leaving a pure salt solution.
(vii)How the ions present in NaCl arrange themselves to form a crystal lattice?
Ans: Na+ and Cl− ions alternate in all three directions. Each ion is surrounded by six oppositely charged ions, giving a cubic lattice.
Na+ Cl− — Fig: NaCl crystal lattice
(viii)Which factors are responsible for the strength of electrostatic forces between ions?
Ans: (1) The magnitude of the charges on the ions, and (2) the distance between the ions (ionic size).
4.Write short answers to any Five (5) questions:[5×2=10]
(i)Is lead (II) chloride soluble in water? Give comments.
Ans: PbCl2 is only sparingly soluble in cold water but dissolves better in hot water. Most chlorides are soluble, except those of silver and lead.
(ii)Which lead salts are insoluble in water?
Ans: Lead sulfate (PbSO4), lead carbonate (PbCO3) and lead chloride (PbCl2, in cold water). Only lead nitrate and lead ethanoate are soluble.
(iii)Name two insoluble carbonates.
Ans: Calcium carbonate (CaCO3) and copper carbonate (CuCO3).
(iv)Why are melting points of ionic salts generally very high?
Ans: Strong electrostatic forces hold the ions in the lattice, so a large amount of energy (heat) is needed to break them apart.
(v)What factors affect the strength of ionic bond.
Ans: (1) Charge on the ions – higher charge gives a stronger bond. (2) Size of the ions – smaller ions (shorter distance) give a stronger bond.
(vi)How would you prepare pure crystals of Barium sulfate in the laboratory?
Ans: By precipitation: mix barium chloride solution with sodium sulfate solution. A white precipitate forms; filter it, wash with distilled water and dry between filter papers.
BaCl2 + Na2SO4 → BaSO4↓ + 2NaCl
(vii)Give an example of salt prepared by titration method.
Ans: Sodium chloride: NaOH + HCl → NaCl + H2O (also sodium sulfate, potassium nitrate).
(viii)Write nuetralization reaction between sulfuric acid and sodium hydroxide.
Ans: H2SO4 + 2NaOH → Na2SO4 + 2H2O
Part – II
NOTE:- Attempt any two questions.[9×2=18]
5.(a) Why do ionic salts exist in solid state?
Ans: In an ionic salt every ion is attracted by strong electrostatic forces to many oppositely charged neighbours. These forces lock the ions into a rigid, regularly repeating crystal lattice. A large amount of energy is required to overcome them, so ionic salts are solids at room temperature, with high melting and boiling points.
(b) Explain the formation of sodium chloride crystal?
Ans:
- Sodium (2,8,1) loses one electron to form Na+ (2,8): Na → Na+ + e−
- Chlorine (2,8,7) gains that electron to form Cl− (2,8,8): Cl + e− → Cl−
- The oppositely charged ions attract each other strongly (ionic bond): Na+ + Cl− → NaCl
- Ions pack alternately in three dimensions; each Na+ is surrounded by six Cl− and vice versa, forming a cubic crystal lattice.
Na+ Cl− — Fig: Sodium chloride crystal lattice
6.(a) How will you prepare calcium sulphate in the laboratory?
Ans: Calcium sulfate is sparingly soluble, so it is prepared by precipitation.
- Take calcium chloride solution in a beaker and add sodium sulfate solution slowly with stirring.
- A white precipitate of CaSO4 forms. CaCl2 + Na2SO4 → CaSO4↓ + 2NaCl
- Filter, wash the residue with distilled water to remove soluble NaCl, then dry between filter papers.
(b) Explain the methods of preparation of two soluble salts.
Ans:
1. Titration method (sodium chloride):
- Pipette 25 cm³ of NaOH into a flask and add 2–3 drops of an indicator (phenolphthalein).
- Add HCl from a burette until the colour just changes (end point); note the volume.
- Repeat without indicator using the same volumes, evaporate to the crystallization point, cool and filter the crystals.
- Warm dilute H2SO4 and add copper(II) oxide in excess with stirring.
- Filter off the unreacted CuO; the blue filtrate is CuSO4 solution.
- Evaporate to the crystallization point, cool to get crystals, then dry.
7.(a) Compare the melting points of NaCl and MgCl2.
Ans: NaCl melts at about 801 °C and MgCl2 at about 714 °C; both are high because of strong electrostatic forces in their lattices.
- Mg2+ carries a double charge and is smaller than Na+, so it attracts Cl− more strongly in principle.
- However, the small, highly charged Mg2+ distorts the Cl− cloud (partial covalent character) and has a different lattice structure, so MgCl2 melts a little lower than NaCl.
(b) How would you prepare pure crystals of copper sulfate in the laboratory?
Ans:
- Take about 25 cm³ of dilute sulfuric acid in a beaker and warm it gently.
- Add copper(II) oxide a little at a time with stirring until no more dissolves (excess).
- Filter to remove excess CuO; collect the blue filtrate.
- Heat the filtrate to evaporate water until the crystallization point (a drop on a glass rod forms crystals on cooling).
- Leave to cool slowly; blue crystals of CuSO4·5H2O form. Filter, wash with a little cold water and dry between filter papers.
