Class 10 Science

Metals and Non-metals

NCERT Exemplar Probelms

Short Answer

Part 1

Question 37: Iqbal treated a lustrous, divalent element M with sodium hydroxide. He observed the formation of bubbles in reaction mixture. He made the same observations when this element was treated with hydrochloric acid. Suggest how can he identify the produced gas. Write chemical equations for both the reactions.

Answer: In the first case, a base reacts with a metal, and in the second case, an acid reacts with a metal. Hence, hydrogen gas is most likely to be produced during these reactions.

The evolution of hydrogen gas can be checked by bringing a burning matchstick or splinter near the evolved gas. The fact that the matchstick burns with a pop sound, shows that hydrogen gas is evolved.

The reactions can be shown by following equations:

2NaOH + Zn ⇨ Na2ZnO2 + H2

2HCl + Zn ⇨ ZnCl2 + H2

Question 38: During extraction of metals, electrolytic refining is used to obtain pure metals.

(a) Which material will be used as anode and cathode for refining of silver metal by this process?

Answer: In electrolytic refining, impure metal is always used as anode and pure metal is used as cathode. Metals are electropositive in nature and hence they are released from anode and get deposited on cathode. Hence, impure silver will be used as anode and pure silver will be used as cathode.

(b) Suggest a suitable electrolyte also.

Answer: Silver chloride

(c) In this electrolytic cell, where do we get pure silver after passing electric current?

Answer: At cathode

Question 39: Why should the metal sulphides and carbonates be converted to metal oxides in the process of extraction of metal from them?

Answer: Obtaining a metal from its oxide is much easier than doing that from a metal sulphide or carbonate. Due to this; metal sulphides and carbonates are first converted to metal oxides during metallurgy.

Question 40: Generally, when metals are treated with mineral acids, hydrogen gas is liberated but when metals (except Mn and Mg), treated with HNO3, hydrogen is not liberated, why?

Answer: Most of the metals do not react with HNO3 because nitric acid is a strong oxidizing agent. Hence, when metals (except Mn and Mg) are treated with nitric acid; hydrogen gas is not liberated.

Question 41: Compound X and aluminium are used to join railway tracks.

(a) Identify the compound X

Answer: Iron oxide (Fe2O3)

(b) Name the reaction

Answer: Thermit reaction

(c) Write down its reaction.

Answer: Fe2O3 + 2Al ⇨ 2Fe + Al2O3 + heat

Question 42: When a metal X is treated with cold water, it gives a basic salt Y with molecular formula XOH (Molecular mass = 40) and liberates a gas Z which easily catches fire. Identify X, Y and Z and also write the reaction involved.

Answer: The salt is NaOH and its molar mass can be calculated as follows:

Na (23) + O(16) + H(1) = 23 + 16 + 1 = 40

It is clear that X is sodium, Y is sodium hydroxide and Z is hydrogen.

The reaction can be shown by following equation:

2Na + 2H2O ⇨ 2NaOH + H2 + Heat

Hydrogen is a highly inflammable gas and hence it easily catches fire.

Question 43: A non-metal X exists in two different forms Y and Z. Y is the hardest natural substance, whereas Z is a good conductor of electricity. Identify X, Y and Z.

Answer: X is carbon, Y is diamond (the hardest natural substance) and Z is graphite (a good conductor of electricity.

Question 44: The following reaction takes place when aluminium powder is heated with MnO2

3MnO2 (s) + 4Al (s) ⇨ 3Mn (l) + 2Al2O3 (l) + Heat

Is aluminium getting reduced?

Answer: Aluminium is getting oxidized.

Is MnO2 getting oxidised?

Answer: MnO2 is getting reduced.

Question 45: What are the constituents of solder alloy? Which property of solder makes it suitable for welding electrical wires?

Answer: Solder is an alloy of lead and tin. The low melting point makes it suitable for welding electrical wires.

Question 46: metal A, which is used in thermite process, when heated with oxygen gives an oxide B, which is amphoteric in nature. Identify A and B. Write down the reactions of oxide B with HCl and NaOH.

Answer: A is aluminium and B is aluminium oxide.

Reaction of aluminium oxide with hydrochloric acid can be written as follows:

Al2O3 + 6HCl ⇨ 2AlCl3 + 3H2O

Reaction of aluminium oxide with sodium hydroxide can be written as follows:

Al2O3 + 2NaOH ⇨ 2NaAlO2 + H2O

Question 47: A metal that exists as a liquid at room temperature is obtained by heating its sulphide in the presence of air. Identify the metal and its ore and give the reaction involved.

Answer: Since the given metal exists as liquid at room temperature, hence it is mercury. Cinnabar (HgS) is an ore of mercury.

The first step of extraction of mercury is heating cinnabar ore in air to get mercury oxide. After that, mercury oxide is reduced to obtain mercury. The following equations show the reactions involved.

2HgS + 3O2 ⇨ 2HgO + 2SO2

2HgO ⇨ 2Hg + O2