Atoms and Molecules

Science

NCERT

1   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 $\to$ sodium ethanoate + carbon dioxide + water

Solution :

In the given reaction, sodium carbonate reacts with ethanoic acid to produce sodium ethanoate, carbon dioxide, and water.$\\$ Sodium carbonate + Ethanoic acid $\to$ Sodium ethanoate + Carbon diaoxide + Water$\\$ $Na_2 CO_3 + CH_3 COOH \to CH_3 COONa + CO_2 + H_2 O$$\\$ Mass of sodium carbonate $= 5.3 g $(Given)$\\$ Mass of ethanoic acid $= 6 g$ (Given)$\\$ Mass of sodium ethanoate $= 8.2 g$ (Given)$\\$ Mass of carbon dioxide $= 2.2 g$ (Given)$\\$ Mass of water $= 0.9 g$ (Given)$\\$ Now, total mass before the reaction $= (5.3 + 6) g = 11.3 g$$\\$ And, total mass after the reaction $= (8.2 + 2.2 + 0.9) g = 11.3 g$$\\$ $\therefore $Total mass before the reaction = Total mass after the reaction$\\$ Hence, the given observations are in agreement with the law of conservation of mass.

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?

Solution :

It is given that the ratio of hydrogen and oxygen by mass to form water is $1:8.$$\\$ Then, the mass of oxygen gas required to react completely with $1 g$ of hydrogen gas is $8 g.$$\\$ Therefore, the mass of oxygen gas required to react completely with $3 g$ of hydrogen gas is$\\$ $8 * 3 g = 24 g.$

3   Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

Solution :

The postulate of Dalton’s atomic theory which is a result of 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?

Solution :

The postulate of Dalton’s atomic theory which can explain the law of definite proportion is: The relative number and kind of atoms in a given compound remains constant.

5   Define atomic mass unit.

Solution :

Mass unit equal to exactly one-twelfth $(\dfrac{1}{12^{th}})$ the mass of one atom of carbon$-12$ is called one atomic mass unit. It is written as $‘u’.$

6   Why is it not possible to see an atom with naked eyes?

Solution :

The size of an atom is so small that it is not possible to see it with naked eyes. Also, the atom of an element does not exist independently.

7   Write down the formulae of$\\$ (i) sodium oxide$\\$ (ii) aluminium chloride$\\$ (iii) sodium suphide$\\$ (iv) magnesium hydroxide

Solution :

(i) Sodium oxide $\to Na _2 O$ $\\$ (ii) Aluminium chloride $\to AlCl_ 3$ $\\$ (iii) Sodium suphide $\to Na _2 S$ $\\$ (iv) Magnesium hydroxide $ \to Mg(OH)_ 2$ $\\$

8   Write down the names of compounds represented by the following formulae:$\\$ (i)$ Al_ 2 (SO_ 4 )_ 3$ $\\$ (ii) $CaCl_ 2$ $\\$ (iii)$ K_ 2 SO_ 4$ $\\$ (iv) $KNO_ 3$ $\\$ (v) $CaCO _3$

Solution :

(i) $Al_ 2 (SO_ 4 )_ 3 \to $ Aluminium sulphate$\\$ (ii)$ CaCl _2 \to $ Calcium chloride$\\$ (iii) $K _2 SO _4 \to $ Potassium sulphate$\\$ (iv) $KNO _3 \to $ Potassium nitrate$\\$ (v) $CaCO _3 \to $ Calcium carbonate

9   What is meant by the term chemical formula?

Solution :

The chemical formula of a compound means the symbolic representation of the composition of a compound. From the chemical formula of a compound, we can know the number and kinds of atoms of different elements that constitute the compound. For example, from the chemical formula $CO _2$ of carbon dioxide, we come to know that one carbon atom and two oxygen atoms are chemically bonded together to form one molecule of the compound, carbon dioxide.

10   How many atoms are present in a$\\$ (i)$ H_ 2 S$ molecule and$\\$ (ii) $PO^{ 3-}_4 $ ion?

Solution :

(i) In an $H_ 2 S$ molecule, three atoms are present; two of hydrogen and one of sulphur.$\\$ (ii) In a $PO ^{3-}_4 $ ion, five atoms are present; one of phosphorus and four of oxygen.

11   Calculate the molecular masses of $H_2, O_2, Cl_2, CO_2, CH_4, C_2H_6, C_2H_4, NH_3, CH_3OH.$ $\\$

Solution :

Molecular mass of $H_2 = 2 ×$ Atomic mass of $H =2×1$$\\$ $=2u$$\\$ Molecular mass of $O_2 = 2 ×$ Atomic mass of$ O = 2 × 16$$\\$ $= 32 u$ Molecular mass of $Cl_2 = 2 × $Atomic mass of $Cl$$\\$ $= 71 u$$\\$ Molecular mass of $CO_2 $= Atomic mass of $C + 2 × $Atomic mass of $O = 12 + 2 × 16$$\\$ $= 44 u$$\\$ Molecular mass of $CH_4$ = Atomic mass of $C + 4 ×$ Atomic mass of $H$ $\\$ $= 12 + 4 × 1$$\\$ $= 16 u$$\\$ Molecular mass of $C_2H_6 =2×$Atomic mass of $C+6×$ Atomic mass of $H$ $\\$ $= 2 × 12 + 6 × 1$ $\\$ $= 30 u$$\\$ Molecular mass of $C_2H_4 =2$×Atomic mass of $C+4×$Atomic mass of $H$ $\\$ $= 2 × 12 + 4 × 1$$\\$ $= 28 u$$\\$ Molecular mass of $NH_3$ = Atomic mass of $N + 3 ×$ Atomic mass of $H$$\\$ $= 14 + 3 × 1$$\\$ $= 17 u$$\\$ Molecular mass of $CH_3OH $= Atomic mass of $C + 4 ×$ Atomic mass of $H $+ Atomic mass of $O$$\\$ $= 12 + 4 × 1 + 16$$\\$ $= 32 u$

12   Calculate the formula unit masses of $ZnO, Na_2O, K_2CO_3$, given atomic masses of $Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, $ and $O = 16 u.$

Solution :

Formula unit mass of $ZnO$ = Atomic mass of $Zn$ + Atomic mass of $O$ $\\$ $= 65 + 16$$\\$ $= 81 u$$\\$ Formula unit mass of $Na2O = 2 ×$ Atomic mass of $Na $+ Atomic mass of $O = 2 × 23 + 16$ $\\$ $= 62 u$ Formula unit mass of $K_2CO_3 =2×$ Atomic mass of $K+$ Atomic mass of $C+3×$Atomic mass of $O$ $\\$ $= 2 × 39 + 12 + 3 × 16$$\\$ $= 138 u$

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

Solution :

One mole of carbon atoms weighs $12 g$ (Given)$\\$ i.e., mass of $1$ mole of carbon atoms = $12 g$$\\$ Then, mass of $6.022*10^{23}$ number of carbon atoms = $12 g$ $\\$ Therefore, mass of $1$ atom of carbon =$\dfrac{ 12} {6.022 *10^{23}}g $$\\$ $=1.9926*10^{-23} g$

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

Solution :

Atomic mass of $Na = 23 u $(Given)$\\$ Then, gram atomic mass of $Na = 23 g$$\\$ Now, $23 g $of $Na$ contains = $6.022*10^{23}$ number of atoms$\\$ Thus, $100 g $of $Na$ contains =$\dfrac{ 6.022*10^{23}}{23} *100$ number of atoms $\\$ $= 2.6182*10^{24 } $number of atoms$\\$ Again, atomic mass of $Fe = 56 u$(Given)$\\$ Then, gram atomic mass of $Fe = 56 g$$\\$ Now,$ 56 g $of $Fe $contains = $6.022*10^{23} $number of atoms$\\$ Thus,$ 100 g$ of $Fe$ contains $= \dfrac{6.022*10^{23}}{56}* 100$ number of atoms $\\$ $=1.0753*10^{24}$ number of atoms$\\$ Therefore, $100 $grams of sodium contain more number of atoms than $100$ grams of iron.

15   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.

Solution :

Mass of boron = $0.096 g$ (Given) $\\$ Mass of oxygen = $0.144 g$ (Given) $\\$ Mass of sample = $0.24 g$ (Given)$\\$ Thus, percentage of boron by weight in the compound $=\dfrac{ 0.096 }{0.24}*100\% $ $\\$ $=40\%$ $\\$ And, percentage of oxygen by weight in the compound $=\dfrac{0.144}{0.24} *100\% $ $\\$

16   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?

Solution :

Carbon + Oxygen $\to$ Carbon dioxide$\\$ $3 g$ of carbon reacts with 8 g of oxygen to produce $11 g$ of carbon dioxide.$\\$ If $3 g$ of carbon is burnt in $50 g$ of oxygen, then $3 g$ of carbon will react with $8 g$ of oxygen. The remaining $42 g$ of oxygen will be left un-reactive.$\\$ In this case also, only $11 g$ of carbon dioxide will be formed.$\\$ The above answer is governed by the law of constant proportions.

17   What are polyatomic ions? Give examples?

Solution :

A polyatomic ion is a group of atoms carrying a charge (positive or negative). For example, ammonium ion $(NH_4^+)$ hydroxide ion $(OH^-)$, carbonate ion , $(CO^{2-}_3)$ sulphate ion $(SO^{2-}_4).$

18   Write the chemical formulae of the following:$\\$(a) Magnesium chloride$\\$ (b) Calcium oxide$\\$ (c) Copper nitrate$\\$ (d) Aluminium chloride $\\$(e) Calcium carbonate

Solution :

(a) Magnesium chloride $\to$$ MgCl_2$ $\\$ (b) Calcium oxide $\to CaO$ $\\$ (c) Copper nitrate $\to Cu (NO_3)_2$ $\\$ (d) Aluminium chloride $\to AlCl_3$ $\\$ (e) Calcium carbonate $\to CaCO_3$ $\\$

19   Give the names of the elements present in the following compounds:$\\$ (a) Quick lime$\\$ (b) Hydrogen bromide$\\$ (c) Baking powder$\\$ (d) Potassium sulphate

Solution :

$\begin{array}{|c|c|} \hline \textbf{Compound } &\textbf{Chemical formula} & \textbf{Elements present}\\ \hline \text{Quick lime }& CAO & \text{Calcium , oxygen}\\ \hline \text{Hydrogen bromide} & HBr & \text{Hydrogen,bromine}\\ \hline \text{Baking powder} & NaHCO_3& \text{Sodium, hydrogen,carbon,oxygen}\\ \hline \text{Potassium sulphate} & k_2SO_4& \text{Potassium, sulphur,oxygen}\\ \hline \end{array}$

20   Calculate the molar mass of the following substances:$\\$ (a) Ethyne, $C_2H_2$ $\\$ (b) Sulphur molecule, $S_8$ $\\$ (c) Phosphorus molecule, $P_4$ (atomic mass of phosphorus = $31$) $\\$(d) Hydrochloric acid, $HCl$ $\\$ (e) Nitric acid, $HNO_3$

Solution :

(a)Molar mass of ethyne, $C_2H_2 =2×12+2×1=28g$ $\\$ (b) Molar mass of sulphur molecule, $S_8 = 8 × 32 = 256 g$ $\\$ (c) Molar mass of phosphorus molecule, $P_4 = 4 × 31 = 124 g $$\\$ (d) Molar mass of hydrochloric acid, $HCl = 1 + 35.5 = 36.5 g $$\\$ (e)Molar mass of nitric acid, $HNO_3 =1+14+3×16=63g$ $\\$

21   What is the mass of:$\\$ (a) $1$ mole of nitrogen atoms?$\\$ (b) $4$ moles of aluminium atoms (Atomic mass of aluminium =$ 27$)? $\\$(c) $10$ moles of sodium sulphite $(Na_2SO_3)$?

Solution :

(a) The mass of $1$ mole of nitrogen atoms is $14 g.$$\\$ (b) The mass of $4$ moles of aluminium atoms is $(4 × 27) g = 108 g $(c) The mass of $10 $moles of sodium sulphite $(Na_2SO_3)$ is$\\$ $10 × [2 × 23 + 32 + 3 × 16] g = 10 × 126 g = 1260 g$

22   Convert into mole.$\\$ (a) $12 g$ of oxygen gas$\\$ (b) $20 g$ of water$\\$ (c) $22 g $ of carbon dioxide

Solution :

(a) $32 g $of oxygen gas = $1$ mole $\\$ Then $12 g$ of oxygen gas = $\dfrac{12}{32}$ mole = $0.375$ mole (b) $18 g$ of water = $1$ mole$\\$ Then, $20 g$ of water = $\dfrac{20}{18}$ mole = $1.11$ moles (approx) $\\$ (c) $44 g$ of carbon dioxide = $1$ mole$\\$ Then, $22 g$ of carbon dioxide = $\dfrac{22}{44} $mole = $0.5$ mole

23   What is the mass of:$\\$ (a) $0.2 $mole of oxygen atoms?$\\$ (b) $0.5$ mole of water molecules?

Solution :

(a) Mass of one mole of oxygen atoms = $16 g$ $\\$ Then, mass of $0.2$ mole of oxygen atoms =$ 0.2 × 16g = 3.2 g$$\\$ (b) Mass of one mole of water molecule = $18 g$$\\$ Then, mass of $0.5$ mole of water molecules = $0.5 × 18 g = 9 g$

24   Calculate the number of molecules of sulphur$ (S_8)$ present in $16 g$ of solid sulphur.

Solution :

$1$ mole of solid sulphur $(S_8) = 8 × 32 g = 256 g$$\\$ i.e., $256 g$ of solid sulphur contains =$6.022 × 10^{23}$ molecules$\\$ Then, $16 g$ of solid sulphur contains $=\dfrac{ 6.022*10^{23} }{256}*16$ molecules $\\$ $= 3.76 × 10^{22}$ molecules (approx)

25   Calculate the number of aluminium ions present in $0.051 g$ of aluminium oxide. (Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of $Al = 27 u$)

Solution :

$1$ mole of aluminium oxide $(Al_2O_3) = 2 × 27 + 3 × 16 = 102 g$ $\\$ i.e., $102 g$ of $Al_2O_3 = 6.022 × 10^{23}$ molecules of $Al_2O_3$ $\\$ Then, $0.051 g$ of $Al_2O_3$ contains = $\dfrac{6.022*10^{23}}{102}*0.051 $ molecules$\\$ $= 3.011 × 10^{20}$ molecules of $Al_2O_3$ $\\$ The number of aluminium ions $(Al^{3+})$ present in one molecule of aluminium oxide is $2.$$\\$Therefore, the number of aluminium ions $(Al^{3+})$ present in $3.011 × 10^{20}$ molecules $(0.051 g )$ of aluminium oxide $(Al_2O_3) = 2 × 3.011 × 10^{20}$ $\\$ $= 6.022 × 10^{20}$