Selina Concise Solutions for ICSE Class 9 Chemistry Chapter 4 The Language Of Chemistry

Exercise 

Question 1.
(a) \( \bigcirc \), \( \bigodot \)
(b) gram
(c) molecular formula
(d) basic, acidic
(e) 4, 3, 2, 1
(f) 2, 3
(g) \( \text{Fe}_2[\text{CO}_3]_3 \)
Answer: (a) symbols \( \bigcirc \), \( \bigodot \)
(b) gram
(c) molecular formula
(d) basic, acidic
(e) 4, 3, 2, 1
(f) 2, 3
(g) \( \text{Fe}_2[\text{CO}_3]_3 \)
In simple words: These are the short answers for various chemistry concepts like identifying old symbols, units of mass, and specific chemical formulas.

📝 Teacher's Note: Use this exercise to test if students can identify transition metal formulas and understand the difference between basic and acidic radicals. Point out that the circles were early symbols used by Dalton.

🎯 Exam Tip: When writing chemical formulas with polyatomic ions like carbonate, always remember to use brackets if the subscript is more than one, like in \( \text{Fe}_2[\text{CO}_3]_3 \).

Question 2.
Answer:

In simple words: This table shows how different positive parts (basic radicals) and negative parts (acid radicals) join together to form specific chemical compounds.

📝 Teacher's Note: This grid is an excellent way to practice the criss-cross method for writing chemical formulas. Ensure students pay attention to the valency of radicals like Phosphate (3-) and Sulphate (2-).

🎯 Exam Tip: Practice writing formulas for polyatomic ions (like Ammonium and Nitrate) frequently, as they are commonly tested in both formula writing and balancing questions.

Question 3.
Answer: Sodium chloride + Silver nitrate \( \rightarrow \) Silver chloride + Sodium nitrate
(a) Equation \( \text{NaCl} + \text{AgNO}_3 \rightarrow \text{AgCl} + \text{NaNO}_3 \)
(b) Yes, the equation is balanced.
(c) \( \text{NaCl} + \text{AgNO}_3 \rightarrow \text{AgCl} + \text{NaNO}_3 \)
\( (23 + 35.5) + (108 + 14 + 48) \rightarrow (108 + 35.5) + (23 + 14 + 48) \)
Wt. of reactants 228.5g      Wt. of products 228.5g
(d) This equation satisfies the "Law of Conservation of Mass."
Law of Conservation of mass: "Matter is neither created nor destroyed in course of a chemical reaction."
In simple words: This experiment proves that when chemicals react, the total weight of the starting materials is exactly the same as the weight of the new products formed.

📝 Teacher's Note: Use the example of a simple addition (\( 228.5 = 228.5 \)) to show students that chemistry obeys the laws of physics. It's a great practical demonstration of the Law of Conservation of Mass.

🎯 Exam Tip: If a question asks to prove the Law of Conservation of Mass, always calculate the total mass of the reactants side and the total mass of the products side separately to show they are equal.

Question 4.
Answer:
(a) \( \text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2 \)
This equation conveys following information:
1. The actual result of chemical change.
2. The substances take part in a chemical reaction and substances formed as a result of reaction.
3. Here one molecule of zinc, one molecule of Sulphuric acid react to give one molecule of zinc sulphate and one molecule of Hydrogen.
4. Composition of respective molecules i.e. one molecule of sulphuric acid contains two atoms of hydrogen, one atom of sulphur and four atoms of oxygen.
5. Relative molecular masses of different substances i.e. molecular mass of \( \text{Zn}= 65, \text{H}_2\text{SO}_4 (2+32+64) = 98, \text{ZnSO}_4 (65+32+64) = 161, \text{H}_2 = 2 \)
6. 22.4 litres of hydrogen are formed at S.T.P.

(b) \( \text{Mg} + 2\text{HCl}_2 \rightarrow \text{MgCl}_2 + \text{H}_2 \)
This equation conveys following information:
1. Magnesium reacts with Hydrochloric acid to form Magnesium chloride and Hydrogen gas.
2. 24g of Magnesium react with \( 2(1 + 35.5) = 73\text{g} \) of Hydrochloric acid to produce \( (24 + 71) \) i.e. 95g of Magnesium chloride
3. That Hydrogen produced out at S.T.P. is 22.4 liters.
In simple words: Chemical equations tell us exactly which ingredients we need, how much they weigh, what we will get as a result, and even the volume of gas produced.

📝 Teacher's Note: Explain that "S.T.P." stands for Standard Temperature and Pressure, where 1 mole of any gas occupies 22.4 liters. This helps bridge the gap between mass and volume in chemistry.

🎯 Exam Tip: When asked what an equation "conveys," mention both qualitative information (names of reactants/products) and quantitative information (masses and gas volumes).

Question 5.
Answer:
(a) A poly-atomic ion is a charged ion composed of two or more atoms covalently bounded that can be carbonate \( (\text{CO}_3^{2-}) \) and sulphate \( (\text{SO}_4^{2-}) \)
(b) The fundamental laws which are involved in every equation are:
1. A chemical equation consists of formulae of reactants connected by plus sign (+) and arrow \( (\rightarrow) \) followed by the formulae of products connected by plus sign (+).
2. The sign of an arrow \( (\rightarrow) \) is to read 'to form'. It also shows the direction in which reaction is predominant.
In simple words: Poly-atomic ions are groups of atoms that act like a single charged unit. Chemical equations use symbols and arrows to tell the story of how chemicals change.

📝 Teacher's Note: Help students visualize poly-atomic ions as "teams" of atoms that usually stay together during a chemical reaction. Use Carbonate and Sulphate as the primary examples.

🎯 Exam Tip: Remember that in a chemical equation, the arrow \( \rightarrow \) always points from the reactants to the products.

Question 6.
Answer: (a) two
(b) six
(c) three
(d) four
(e) (i) three (ii) five (iii) four (iv) two
In simple words: These are counts of atoms or molecules in specific chemical setups, often used when figuring out how to balance an equation.

📝 Teacher's Note: This exercise likely refers to counting atoms in given formulas. Practice counting atom subscripts inside and outside brackets to improve accuracy.

🎯 Exam Tip: For sub-part (e), ensure you understand how coefficients multiply through the entire chemical formula.

Question 7.
Answer: According to law of conservation of mass, "matter can neither be created nor be destroyed in a chemical reaction". This is possible only, if total number of atoms on the reactants side is equals to total number of atoms on products side. Thus, a chemical reaction should be always balanced.
Let us consider an example,
\( \text{Fe} + \text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + \text{H}_2 \)
In this equation number of atoms on both sides is not the same, the equation is not balanced.
The balanced form of this equation is given by,
\( 3\text{Fe} + 4\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + 4\text{H}_2 \)
In simple words: We balance equations because atoms don't just disappear or appear out of nowhere. We need the same number of "LEGO bricks" of each element on both sides of the reaction.

📝 Teacher's Note: Introduce balancing as a puzzle. Students must adjust coefficients (big numbers in front) but never change the subscripts (small numbers) within a formula.

🎯 Exam Tip: Always double-check your atom count for each element after you think you've finished balancing an equation.

Question 8.
Answer:
(a) \( 3\text{Fe} + 4\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + 4\text{H}_2 \)
(b) \( 3\text{Ca} + \text{N}_2 \rightarrow \text{Ca}_3\text{N}_2 \)
(c) \( \text{Zn} + 2\text{KOH} \rightarrow \text{K}_2\text{ZnO}_2 + \text{H}_2 \)
(d) \( \text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2 \)
(e) \( 3\text{PbO} + 2\text{NH}_3 \rightarrow 3\text{Pb} + 3\text{H}_2\text{O} + \text{N}_2 \)
(f) \( 2\text{Pb}_3\text{O}_4 \rightarrow 6\text{PbO} + \text{O}_2 \)
(g) \( 2\text{PbS} + 3\text{O}_2 \rightarrow 2\text{PbO} + 2\text{SO}_2 \)
(h) \( \text{S} + 2\text{H}_2\text{SO}_4 \rightarrow 3\text{SO}_2 + 2\text{H}_2\text{O} \)
(i) \( \text{S} + 6\text{HNO}_3 \rightarrow \text{H}_2\text{SO}_4 + 6\text{NO}_2 + 2\text{H}_2\text{O} \)
(j) \( \text{MnO}_2 + 4\text{HCl} \rightarrow \text{MnCl}_2 + 2\text{H}_2\text{O} + \text{Cl}_2 \)
(k) \( \text{C} + 2\text{H}_2\text{SO}_4 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} + 2\text{SO}_2 \)
(l) \( 6\text{KOH} + 3\text{Cl}_2 \rightarrow 5\text{KCl} + \text{KClO} + 3\text{H}_2\text{O} \)
(m) \( 2\text{NO}_2 + \text{H}_2\text{O} \rightarrow \text{HNO}_2 + \text{HNO}_3 \)
(n) \( \text{Pb}_3\text{O}_4 + 8\text{HCl} \rightarrow 3\text{PbCl}_2 + 4\text{H}_2\text{O} + \text{Cl}_2 \)
(o) \( 2\text{H}_2\text{O} + 2\text{Cl}_2 + \text{Sunlight} \rightarrow 4\text{HCl} + \text{O}_2 \)
(p) \( 2\text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2 \)
(q) \( 2\text{HNO}_3 + \text{H}_2\text{S} \rightarrow 2\text{NO}_2 + 2\text{H}_2\text{O} + \text{S} \)
(r) \( \text{P} + 5\text{HNO}_3 \rightarrow 5\text{NO}_2 + \text{H}_2\text{O} + \text{H}_3\text{PO}_4 \)
In simple words: This is a long list of chemical reactions that have been balanced so that the number of atoms for every element is equal on both the left and right sides.

📝 Teacher's Note: This is a comprehensive list for balancing practice. Focus on equations (e) and (n) as they are more complex and require careful atom tracking.

🎯 Exam Tip: When balancing redox-like reactions such as in (i) or (q), start by balancing the elements that appear in only one compound on each side first.

Question 9.
Answer:
(a) \( 2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \)
(b) \( 2\text{KHCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{CO}_2 + 2\text{H}_2\text{O} \)
(c) \( \text{Fe} + \text{H}_2\text{SO}_4 \rightarrow \text{Fe(SO}_4) + \text{H}_2 \)
(d) \( \text{Cl}_2 + \text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4 + 2\text{HCl} \)
(e) \( 2\text{AgNO}_3 \rightarrow 2\text{Ag} + 2\text{NO}_2 + \text{O}_2 \)
(f) \( 3\text{Cu} + 8\text{HNO}_3 \rightarrow 3\text{Cu(NO}_3)_2 + 2\text{NO} + 4\text{H}_2\text{O} \)
(g) \( 4\text{NH}_3 + 5\text{O}_2 \xrightarrow{\text{Pt, } 800^\circ\text{C}} 6\text{H}_2\text{O} + 4\text{NO} + \text{Heat} \)
(h) \( \text{BaCl}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2\text{HCl} \)
(i) \( 2\text{ZnS} + 3\text{O}_2 \rightarrow 2\text{ZnO} + 2\text{SO}_2 \)
(j) \( \text{Al}_4\text{C}_3 + 12\text{H}_2\text{O} \rightarrow 4\text{Al(OH)}_3 + 3\text{CH}_4 \)
(k) \( 4\text{FeS}_2 + 11\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 + 8\text{SO}_2 \)
(l) \( 2\text{KMnO}_4 + \text{HCl} \rightarrow 2\text{KCl} + 2\text{MnCl}_2 + 5\text{Cl}_2 + 8\text{H}_2\text{O} \)
(m) \( \text{Al}_2\text{(SO}_4)_3 + 8\text{NaOH} \rightarrow 3\text{Na}_2\text{SO}_4 + 2\text{NaAlO}_2 + 4\text{H}_2\text{O} \)
(n) \( 2\text{Al} + 2\text{NaOH} + 2\text{H}_2\text{O} \rightarrow 2\text{NaAlO}_2 + 3\text{H}_2 \)
(o) \( 2\text{K}_2\text{Cr}_2\text{O}_7 + 8\text{H}_2\text{SO}_4 \rightarrow 2\text{K}_2\text{SO}_4 + 2\text{Cr}_2\text{(SO}_4)_3 + 8\text{H}_2\text{O} + 3\text{O}_2 \)
(p) \( \text{K}_2\text{Cr}_2\text{O}_7 + 14\text{HCl} \rightarrow 2\text{KCl} + 2\text{CrCl}_3 + 7\text{H}_2\text{O} + 3\text{Cl}_2 \)
(q) \( \text{S} + 6\text{HNO}_3 \rightarrow \text{H}_2\text{SO}_4 + 6\text{NO}_2 + 2\text{H}_2\text{O} \)
(r) \( 2\text{KI} + 2\text{MnO}_2 + 4\text{H}_2\text{SO}_4 \rightarrow \text{I}_2 + 2\text{KHSO}_4 + 2\text{MnSO}_4 + 4\text{H}_2\text{O} \)
In simple words: This is another list of balanced equations, showing many different types of chemical reactions, including some that need special heat or catalysts to work.

📝 Teacher's Note: Note equation (g); it's the Ostwald process for making nitric acid. Use it to explain how catalysts like Platinum (Pt) speed up reactions without being consumed.

🎯 Exam Tip: Pay attention to equations involving Potassium Dichromate \( (\text{K}_2\text{Cr}_2\text{O}_7) \), as they often appear in higher-level exams and involve many molecules of \( \text{HCl} \) or \( \text{H}_2\text{SO}_4 \).

Question 10.
Answer: (a) The atomic mass unit (amu) is defined as \( 1/12\text{th} \) of the mass of an atom of carbon.
1 a.m.u. = \( 1.67 \times 10^{-24}\text{ gm} = 1.67 \times 10^{-27}\text{ kg} \)
1 gm mass = \( 6.02 \times 10^{23}\text{ a.m.u.} \) and 1 kg mass = \( 6.02 \times 10^{26}\text{ a.m.u.} \)

(b)
(i) The relative molecular mass of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \)
= \( 63.5 + 32 + (16 \times 4) + 5 (2 + 16) \)
= \( 159.5 + 90 = 249.5 \)

(ii) The relative molecular mass of \( \text{(NH}_4)_2\text{CO}_3 = \text{N}_2\text{H}_8\text{CO}_3 \)
= \( 14 \times 2 + 1 \times 8 + 12 + 3 \times 16 \)
= \( 28 + 8 + 12 + 48 = 96 \)

(iii) The relative molecular mass of \( \text{(NH}_2)_2\text{CO} = \text{N}_2\text{H}_4\text{CO} \)
= \( 2 \times 14 + 1 \times 4 + 12 + 16 \)
= \( 28 + 4 + 12 + 16 = 60 \)

(iv) The relative molecular mass of \( \text{Mg}_3\text{N}_2 \)
= \( 3 \times 24 + 2 \times 14 = 72 + 28 = 100 \)
In simple words: Atomic mass unit is a tiny scale for weighing atoms. Molecular mass is just adding up the weights of all the individual atoms that make up a molecule.

📝 Teacher's Note: When calculating molecular mass for hydrates like \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \), remind students that the dot means they should add the mass of the 5 water molecules, not multiply by them.

🎯 Exam Tip: Always show your step-by-step addition for molecular mass calculations to ensure you don't make small arithmetic errors that lose marks.

Question 11.
(a) (i) Dalton (ii) Proust (iii) Berzelius (iv) Lavoisier
(b) (i) One (ii) Two (iii) Three (iv) Four
(c) (i) \( \text{FeSO}_4 \) (ii) \( \text{Fe}_2\text{SO}_4 \) (iii) \( \text{Fe}_2\text{(SO}_4)_3 \) (iv) \( \text{Fe}_3\text{(SO}_4)_2 \)
(d) (i) 1: 8 (ii) 8: 1 (iii) 1: 2 (iv) 2: 1
(e) (i) \( \text{CaHCO}_3 \) (ii) \( \text{Ca(HCO}_3)_2 \) (iii) \( \text{Ca}_2\text{HCO}_3 \) (iv) \( \text{Ca(HCO}_3)_3 \)
Answer: (a) (iii) Berzelius
(b) (i) One
(c) (iii) \( \text{Fe}_2\text{(SO}_4)_3 \)
(d) (i) 1: 8
(e) (ii) \( \text{Ca(HCO}_3)_2 \)
In simple words: These are correct answers to multiple-choice questions about chemistry history, atom counts, and chemical formulas.

📝 Teacher's Note: Point out that the ratio of 1:8 in water (\( \text{H}_2\text{O} \)) refers to mass (\( 2 \text{ grams of H to } 16 \text{ grams of O} \)), which simplifies to 1:8.

🎯 Exam Tip: For valency-based formula questions like (c) and (e), use the valency of the metal (Iron III is 3+) and the radical (Sulphate is 2-) to verify the subscripts.

Question 12.
Answer: (a) A molecular formula represent The Molecule of an element or of a Compound.
(b) The molecular formula of water \( (\text{H}_2\text{O}) \) represents 18 parts by mass of water.
(c) A balanced equation obeys the law of conservation of mass wherever unbalanced equation does not obey this law.
(d) CO and Co represent carbon-monoxide and cobalt respectively.
In simple words: Formulas show us exactly what a substance is made of and how much it weighs. We must use correct capitalization (like CO vs Co) because it changes the whole meaning!

📝 Teacher's Note: Emphasize sub-part (d) very strongly. Students often confuse elements (Co) with compounds (CO). Using capital letters correctly is vital in chemical notation.

🎯 Exam Tip: When defining a molecular formula, mention that it indicates both the types of atoms present and the actual number of each atom in one molecule.