NCERT Solutions for Science Class 9 Chapter 3

Chapter 3: Atoms and Molecules

Intext Exercise 1

Question: In a reaction, 5.3g of sodium carbonate reacted with 6g of ethanoic acid. The products were 2.2g of carbon dioxide, 0.9g water and 8.2g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass. 

Sodium Carbonate + ethanoic acid → sodium rthanoate + carbon dioxide + water

Answer: 

Given,

Mass of sodium carbonate \(=5.3\space g\)

Mass of ethanoic acid \(=6\space g\)

Mass of carbon rthanoate \(=8.2\space g\)

Mass of carbon dioxide \(=2.2\space g\)

Mass of water \(=0.9\space g\)

Now, total mass after the reaction \(=(5.3+6)\space g=11.3\space g\)

And, total mass after the reaction \(=(8.2+2.2+0.9)\space g=11.3\space g\)

Therefore, total mass befor the reaction = Total mass after the reaction

Hence, this is in agreement with the law of conservation of mass.

Question 2: Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?

Answer: In water, \(H:O\) (by mass) \(=1:8\)

The mass of oxygen gas required to react completely with \(1\space g\) of hydrogen gas \(=8\space g\)

So, the mass of oxygen gas required to react completely with \(3\space g\) of hydrogen gas \((=8\times3)g=24\space g\)

Question 3: 3. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Answer: The postulate of Dalton’s atomic theory which is based on the law of conservation of mass is: “Atoms are indivisible particles, which can neither be created nor destroyed in a chemical reaction.”
4. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Answer: “The elements consist of atoms having fixed mass and that the number and kind of atoms of each element in a given compound is fixed.” This explains the law of definite proportion. 

Exercise 2: 

Question 1: Define atomic mas unit.

Answer: Mass unit equal to exactly one-twelfth the \((\frac{1}{12^{th}})\) mass of one atom of carbon \(-12\) is called one atomic mass unit. It is represented ad ‘a.m.u.’ or ‘u’.

Question 2: Why is it not possible to see an atom with naked eyes?

Answer: Due to the small size of an atom we cannot see them with naked eyes.

Exercise 3:
1. Write down the formulae of
(i) Sodium oxide

Answer: Sodium Oxide \(Na_2O\)

(ii) Aluminum chloride 

Answer: Aluminum chloride: \(AlCl_3\)

(iii) Sodium sulphide

Answer: Sodium sulphide: \(Na_2S\)

(iv) Magnesium hydroxide

Answer: Magnesium hydroxide: \(Mg(OH)_2\)

Question 2: Write down the names of compounds represented by the following formula:

(i) \(Al_2(SO_4)_3\)

Answer: Aluminium Sulphate

(ii) \(CaCl_2\)

Answer: Calcium chloride

(iii) \(K_2SO_4\)

Answer: Potassium shlphate

(iv) \(KNO_3\)

Answer: Potassium nitrate

(v) \(CaCO_3\)

Answer: Calcium carbonate

Question 3: What is meant by the term chemical formula?

Answer: The symbolic representation of comosition of a compound is known as chemical formula. Chemical formula gives us the idea of the number of atoms present.

Example: From the chemical formula \(CO_2\) of Carbon Dioxide, we come to know that one carbon atom and two oxygens atoms are chemically bonded together to form one molecule of the compound, carbon dioxide. 

(i) \(H_2S\) molecule

Answer: There are total \(3\) atoms present in \(H_2S\) molecule, two hydrogen atoms and one Sulphur atom.

(ii) \(PO^{3-}_4\) ion

Answer: There are a total \(5\) atoms in \(PO^{3-}_4\) ion, one phosphorus atom and \(4\) oxygen atoms.

Exercise 4: 

Question 1: Calculate the molecular masses of \(H_2,O_2,Cl_2,CH_4,C_2H_2,C_2H_6,C_2H_4,NH_3<CH_3,OH\).

Answer: Molecular mass of \(H_2=2\times\) Atomic mass of \(H\)

\(=2\times1\space u=2\space u\)

\(<plecular mass of \(O_2=2\times\) Atomic of mass of \(O\)

\(=2\times16\space u=32\space u\)

Molecular mass of \(Cl_2=2\times\) Atomic mass of \(Cl\)

\(=2\times35.5\space u=71\space u\)

Molecular mass of \(CO_2=\) Atomic mass of \(C+2\times\) Atomic mass of \(O\)

\(=(12+2\times16)u=44\space u\)

Molecular mass of \(CH_4=\) Atomic mass of \(C+4\times\) Atomic mass of \(H\)

\(=(12+4\times1)\space u=16\space u\)

Molecular mass of \(C_2H_6=2\times\) Atomic mass of \(C+6\times\) Atomic mass of \(H\)

\(=(2\times12+6\times1)\space u=30\space u\)

Molecular mass of \(C_2H_2=2\times\) Atomic mass of \(C+4\times\) Atomic mass of \(H\)

\(=(2\times12+6\times1)u=30\space u\)

Molecular mass of \(NH_3=\) Atomic mass of \(N+3\times\) Atomic mass of \(H\)

\(=(14+3\times1)\space u=17\space u\)

Molecular mass of \(CH_3OH=\) Atomic mass of \(C+3\times\) Atomic mass of \(H+\) Atomic mass of \(O+\) Atomic mass of \(=(12+\times4+16)u=32\space u\)

Question 2: Calculate the fomula unit masses of \(ZnO,\space Na_2,\space O,\space K_2\space CO_3\) given atomic masses of \(Z=65u,\space Na=23u,\space K39u,\space C=12u\) and \(O=16u\).

Answer: Formula unit mass of \(Zno=\) Atomic mass of \(Zn+\) Atomic mass of O=(65+16)u=8\space 81\space u\)

Formula unit mass of \(Na_2,O=2\times\) Atomic mass of \(Na+\) Atomic mass of \(O=(2\times23+16)u=62u\)

Formula unit mass of \(K_2CO_3=2\times\) Atomic mass of \(K+\) Atomic mass of \9C+3\times\) Atomic mass \(O=(2\times39+12+3\times16)u=138\space u\)

Question 1: If one mole of carbon atoms weighs 12 grams, what is the mass (in gram) of 1 atom of carbon? 

Answer: Given mass of one mole of carbon atoms \(=12\space g\)

Therefore, Mass of \(6.022\times10^{23}\), number of carbon atoms \(=12\space g\)

Mass of \(1\) atom of carbon will be:

\(=\frac{12}{6.022\times10^{23}}g\)

\(=1.9927\times10^{-23}\space g\)

Question 2: Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na = 23 u, Fe = 56 u)?

Answer: Atomic mass of \(Na=23\space u\) (Given)

Then, gram atomic mass of \(Na=23\space g\)

Now, \(23\space g\) of \(Na\) contains \(=6.022\times10^{23}\)

Number of \(Na\) atoms

Thus, \(100\space g\) of \(Na\) contains \(=\frac{6.022\times10^{23}}{23}\times100\)

Number of \(Na\) atoms \(=2.6182\times10^{24}\) number of \(Na\) atoms

Atomic mass of \(Fe=56\space u\) (Given)

Then, gram atomic mass of \9Fe=56\space g\)

Now, \(56\space g\) of \9Fe\) contains \(=6.022\times^{23}\) number of \(Fe\) atoms

Thus, \(56\space g\) of \9Fe\) contains \(=\frac{6.022\times10^{23}}{56{\times100\)

Number of \9Fe\) atoms \(=1.0753\times10^{24}\) Number of \(Fe\) atoms

\(2.6182\times10^{24}>1.0753\times10^{24}\)

Therefore, \(100\) grams of sodium contain a greater number of atoms than \(100\) grams of iron.

NCERT Question:

Question 1: A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.

Answer: 

Given,
Mass of boron = 0.096 g
Mass of oxygen = 0.144 g
Mass of sample = 0.24 g
The percentage of boron by weight in the compound \(=\frac{0.096}{0.24}\times100%=40%\)

And, percentage of oxygen by weight in the compound \(=\frac{0.144}{024}\times100%=60%\)

Question :When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combinations will govern your answer?

Answer: Carbon + Oxygen → Carbon dioxide

\(3\space g\) of carbon reacts with \(8\space g\) of oxygen to produce \(11\space g\) of carbon dioxide.

If \(3\space g\) of carbon is burnt in \(50\space g\) of oxygen, then \(3\space g\) of carbon will rect with \(8\space g\) of oxygen to form \(11\space g\) of carbon dioxide.

The remaining \((50-8)=42\space g\) of oxygen will be left unreacted.

The above anser is governed by the law of constant proportions.

Question 3: What are polyatomic ions? Give examples?

Answer: A polyatomic ions is group of atoms carrying a charge either positive or negative.

For example

Ammonium ion \((NH^+_4)\), hydroxide ion \((OH^-)\), carbonate ion, \((CO^{2-}_3)\), sulphate ion \((SO^{2-}_4)\)

Question 4: Qrite the chemical formula of the following:

(a) Magnesium chloride

Answer: \(MgCl_2\)

(b) Calcium oxide

Answer: \(CaO\)

(c) Copper nitrate

Answer: \(Cu(NO_3)_2\)

(d) Aluminium chloride

Answer: \(AlCl_3\)

(e) Calcium carbonate

Answer: \(CaCO_3\)

Question 5: Give the names of the elements present in the following compounds:

(a) Quick lime

(b) Hydrogen bromide

(c) Baking powder

(d) Potassium sulphate

Answer: 

Question 6: Calculate the molar mass of the following substances: 

(a) Ethyne \(C_2H_2\)

Answer: Molar mass of 

\(C_H_2=2\times12+2\times1=38g/mol\)

(b) Sulphur molecule, \(S_8\)

Answer: Molar mass of \(S_8=8\times32=256g/mol\)

(c) Phophorus molecule \(P_4\)

(atomic mass of phosphorus =31)

Answer: Molar mass of \(P_4=4\times31=124g/mol\)

(d) Hydrochloric acid \(HCl\)

Answer: Molar mass of \(HCl=1+35.5=36.5g/mol\)

(e) Nitric acid, \(HNO_3\)

Answer: Molar mass of \(HNO_3=1+14+3\times16=64g/mol\)

Question 7: What is the mass of:

(a) 1 mole of Nitrogen atoms?

Answer: The mass of 1 mole of \(N-\) atoms \(=14\space g\)

(b) 4 moles of Aluminum atoms 9Atomic mass of aluminum =27)?

Answer:

[Atomic mass of \(Al=27u\)]

The mass of 4 moles of \(Al\)- atoms \(=94\times27)g=108\space g\)

(c) 10 moles of Sodium sulphite \((Na_2,SO_3)\)?

Answer: Atomic mass of \(Na=23\space u\), Atomic mass of \(S=32\spce u\), Atomic mass of \(o=16\space u\)

The mass of 10 moles of sodium sulphite.

\((Na_2SO_3)=10\times[2\times23+32+3\times16]g=10\times126g=1260\space g\)