NCERT Exemplar Solutions
Class 10 - Science - CHAPTER 2: Acids, Bases and Salts
Long Answer Questions

Question. 43

43. In the following schematic diagram for the preparation of hydrogen gas as shown in Figure 2.3, what would happen if following changes are made?

(a) In place of zinc granules, same amount of zinc dust is taken in the test tube

(b) Instead of dilute sulphuric acid, dilute hydrochloric acid is taken

(c) In place of zinc, copper turnings are taken

(d) Sodium hydroxide is taken in place of dilute sulphuric acid and the tube is heated

Answer:

(a) Hydrogen gas will evolve with greater speed (because powder has larger surface area).
(b) Almost same amount of gas is evolved (HCl is also a strong acid; Zn + 2HCl → ZnCl₂ + H₂).
(c) Hydrogen gas is not evolved (copper is less reactive than hydrogen and does not react with dilute acid under normal conditions).
(d) If NaOH (a strong base) is used with zinc and heated, hydrogen gas is evolved and sodium zincate is formed: \(\mathrm{Zn + 2NaOH \xrightarrow{\Delta} Na_2ZnO_2 + H_2}\).

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Question. 44

44. For making cake, baking powder is taken. If at home your mother uses baking soda instead of baking powder in cake,

(a) how will it affect the taste of the cake and why?

(b) how can baking soda be converted into baking powder?

(c) what is the role of tartaric acid added to baking soda?

Answer:

(a) The cake will taste bitter because heating NaHCO₃ gives Na₂CO₃ (sodium carbonate) which is bitter: \(\mathrm{2NaHCO_3 \xrightarrow{heat} Na_2CO_3 + H_2O + CO_2}\).
(b) Convert baking soda to baking powder by adding an appropriate amount of tartaric acid (or another edible acid) to produce a mixture that evolves CO₂ on heating.
(c) Tartaric acid neutralises some of the sodium carbonate formed, preventing bitterness and providing controlled CO₂ release for leavening.

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Question. 45

45. A metal carbonate X on reacting with an acid gives a gas which when passed through a solution Y gives the carbonate back. On the other hand, a gas G that is obtained at anode during electrolysis of brine is passed on dry Y, it gives a compound Z, used for disinfecting drinking water. Identify X, Y, G and Z.

Answer:

X = calcium carbonate (\(\mathrm{CaCO_3}\)).
Gas from acid + carbonate = \(\mathrm{CO_2}\).
Y is lime water (\(\mathrm{Ca(OH)_2}\)) which on passing CO₂ gives back calcium carbonate: \(\mathrm{Ca(OH)_2 + CO_2 \to CaCO_3 + H_2O}\).
G (gas at anode in electrolysis of brine) = chlorine (\(\mathrm{Cl_2}\)).
Z (product when dry Cl₂ is passed on dry Ca(OH)₂) = bleaching powder (calcium oxychloride, \(\mathrm{CaOCl_2}\)).

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Question. 46

46. A dry pellet of a common base B, when kept in open absorbs moisture and turns sticky. The compound is also a by-product of chloralkali process. Identify B. What type of reaction occurs when B is treated with an acidic oxide? Write a balanced chemical equation for one such reaction.

Answer:

B = sodium hydroxide (\(\mathrm{NaOH}\)).
Type of reaction with acidic oxide (e.g. CO₂) — acid–base reaction (neutralisation) producing a salt and water/other product. Example: \(\mathrm{2NaOH + CO_2 \to Na_2CO_3 + H_2O}\).

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Question. 47

47. A sulphate salt of Group 2 element of the Periodic Table is a white, soft substance, which can be moulded into different shapes by making its dough. When this compound is left in open for some time, it becomes a solid mass and cannot be used for moulding purposes. Identify the sulphate salt and why does it show such a behaviour? Give the reaction involved.

Answer:

The salt is Plaster of Paris (calcium sulphate hemihydrate, \(\mathrm{CaSO_4\cdot\tfrac{1}{2}H_2O}\)). On absorbing moisture it converts to gypsum (calcium sulphate dihydrate, \(\mathrm{CaSO_4\cdot2H_2O}\)) which is hard. Reaction: \(\mathrm{CaSO_4\cdot\tfrac{1}{2}H_2O + \tfrac{3}{2}H_2O \to CaSO_4\cdot2H_2O}\) or written as \(\mathrm{CaSO_4\cdot\tfrac{1}{2}H_2O + H_2O \to CaSO_4\cdot2H_2O}\) in stoichiometric form.

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Question. 48

48. Identify the compound X on the basis of the reactions given below. Also, write the name and chemical formulae of A, B and C.

Compound X + Zn → A + H₂(g)

Compound X + HCl → B + H₂O

Compound X + CH₃COOH → C + H₂O

Answer:

X = sodium hydroxide (NaOH).
A = sodium zincate, \(\mathrm{Na_2ZnO_2}\) (formed when Zn reacts with hot concentrated NaOH).
B = sodium chloride (NaCl).
C = sodium acetate, \(\mathrm{CH_3COONa}\).

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NCERT Exemplar Solutions Class 10 – Science – CHAPTER 2: Acids, Bases and Salts – Long Answer Questions | Detailed Answers