Atoms And Molecules

Let the mass of quicklime formed is “x”

According to the law of conservation of mass, the sum of the masses of reactants is equal to the sum of the masses of products formed.

Hence, 100g = x + 44g

⇒ x = 100 – 44g

⇒ x = 56 g

Law of conservation of mass was followed for solving this problem.

Law of constant proportion or law of definite proportion states that in a chemical compound, elements always combine in a fixed proportion.

There will be no change in the mass after the reaction.

According to the law of conservation of mass which states that atoms are neither created nor destroyed in a chemical reaction. This means that the total mass of the products formed in chemical reaction must be equal to the mass of reactants consumed.

In the first sample:

9g of hydrogen and 42g of nitrogen are present. According to the law of definite proportion, hydrogen and nitrogen are present in the ratio of 9:42 i.e., 3:14

In the second sample:

It is given that 5g of hydrogen is present along with nitrogen.

According to the law of constant proportion the second sample

must also contain hydrogen and nitrogen in 3:14 ratio.

⇒

⇒ 14 × 5 = 3 × amt. of nitrogen

⇒ Amount of nitrogen = 70/3

⇒ Amount of nitrogen = 23.3g

Thus, the amount of nitrogen in the second sample is 23.3g.

Atoms contains protons, neutrons, and electrons. Dalton was not able to explain this concept. According to him, atoms are indivisible particles. Hence, the definition of an atom given by Dalton is no longer valid.

It is necessary to use the symbol for the elements because by writing symbols of the elements, chemical equations become easier to write as well as to understand and saves time too.

For example:

Zinc – Zn

Hydrogen – H

Aluminium – Al

Calcium – Ca

Sodium – Na

Lithium – Li

(i) Correct symbol for sodium is Na. The symbol of sodium was derived from its latin name “natrium”.

(ii) Correct name for hydrogen is “H”. The symbol “Hg” is used for mercury.

(iii) Correct symbol for copper is Cu. The symbol “Co” is used for cobalt.

(iv) The given symbol of sulphur (S) is correct.

(v) Correct symbol for calcium is “Ca”. According to IUPAC, the first letter of the symbol is always written as a capital letter (uppercase) and the second letter is always written as a small letter (lowercase). Both letters cannot be written as capital letters.

Elements whose symbols do not start with the same letter as that of the name of the element are:

The symbols of a few elements like sodium do not start with the initial letter of the name because their symbol was derived from their names in Latin.

For example:

i. The symbol of sodium (Na) was derived from its latin name “natrium”

ii. The symbol of potassium (K) was derived from its latin name “kalium”

iii. The symbol of iron (Fe) was derived from its latin name “ferrum”

The difference between 2Cl and Cl₂:

i. 2Cl indicates the two chloride ions.

ii. Cl₂ indicates one chlorine molecule which is formed by combining of two atoms of chlorine.

iii. Cl₂ is also known as chlorine gas.

iv. Cl₂ form exists in nature. The atomic mass of this element is

(Atomic mass of chlorines × 2) = 35.5 × 2 = 71u

Atomicity: The number of atoms constituting a molecule is known as its atomicity.

i. In I₂, there are two atoms of iodine constituting a molecule.

Hence, the atomicity is 2 (diatomic).

ii. In H₂S, there are two atoms of hydrogen and one atom of Sulphur constituting a molecule of H₂S. Hence, the atomicity is 3 (triatomic).

iii. In HNO₃, there are three atoms of oxygen, one atom of nitrogen and one atom of hydrogen sulphur constituting a molecule of HNO₃. Hence, the atomicity is five.

iv. In Na₂SO₄, there are two atoms of sodium, one atom of Sulphur and four atoms of oxygen constituting a molecule of Na₂SO₄ Hence, the atomicity is seven.

v. In S₈, there are eight atoms of sulphur constituting a molecule.

Hence, the atomicity is eight.

It is given that nitrogen and oxygen are combined together in the ratio of 7:16 by mass.

Mass of nitrogen = 14u

Mass of oxygen = 16u and mass of O₂ = 16× 2 = 32u

Hence, we can write 14:32 which is equal to 7:16

Therefore, the formula of the oxide is NO₂ and its name is nitrous dioxide.

Difference between ion and atom:

Difference between cation and anion:

i. Calcium phosphate [Ca₃(PO₄)₂] :

There are three calcium atoms, two phosphorus atoms and eight oxygen atoms.

Total number of atoms = 2Ca + 2P + 8O

Hence, there are total 12 atoms.

ii. Hydrogen sulphide (H₂S) :

There are two hydrogen atoms and one sulphur atom.

Total number of atoms = 2H + 1S

Hence, there are total 3 atoms.

iii. Magnesium bromide (MgBr₂) :

There are two bromine atoms and one magnesium atom.

Total number of atoms = 2Br + 1Mg

Hence, there are total 3 atoms.

iv. Sodium oxide (Na₂O):

There are two sodium atoms and one oxygen atom.

Total number of atoms = 2Na + 1O

Hence, there are total 3 atoms.

v. Aluminium hydroxide [Al(OH)₃]:

There are three hydrogen and oxygen atoms and one atom of aluminum.

Total number of atoms = 3H + 3O + 1Al

Hence, there are total 7 atoms.

Hydroxide ion:

Symbol of hydroxide ion = OH^-

Valency = -1

Carbonate ion:

Symbol of carbonate ion = CO₃^2-

Valency = -2

Role of valency:

i. The combining power of an element is known as its valency.

ii. Valency can be used to find out how the atoms of an element will combine with the atoms of another element to form a compound.

X₂(CO₃)₃

From this, we came to know that valency of X is 3+

The formulafor the phosphate of the element X:

Formula: X₃(PO₄)₃ or XPO₄

The formulafor the chloride of the element X:

Formula: XCl₃

i. Formula of magnesium sulphate:

Formula: MgSO₄

ii. Formula of sodium bromide:

Formula: NaBr

iii. The formula of calcium chloride:

Formula: CaCl₂

iv. Formula of potassium nitrate:

Formula: KNO₃

v. Formula of sodium phosphate:

Formula: Na₃PO₄

Molar mass is the mass of one mole of substance, while molecular mass is the mass of one molecule of the substance. Molecular mass is the larger quantity because it contains the mass of one molecule (made up of atoms).

i. MgO:

Molar mass of Mg + 2 × Molar mass of oxygen

⇒ 24u + 2×16

⇒ 24 + 32

⇒ 56g/mol
Thus, the molar mass of MgO is 56g/mol.

ii. Na₂CO₃:

2 × Molar mass of Na + Molar mass of carbon + 3×molar mass of oxygen

⇒ 2 × 23 + 12 + 3 × 16

⇒ 46 + 12 + 48

⇒ 106g/mol

Thus, the molar mass of Na₂CO₃ is 106g/mol.

iii. H₃PO₄:

3 × Molar mass of hydrogen + Molar mass of phosphorus + 4×molar mass of oxygen

⇒ 3 × 1 + 31 + 4 × 16

⇒ 3 + 31 + 64

⇒ 98g/mol

Thus, the molar mass of H₃PO₄ is 98g/mol.

iv. Ca(OH)₂

Molar mass of calcium + 2 × Molar mass of oxygen + 2×molar mass of hydrogen

⇒ 40 + 2 × 16 + 2 × 1

⇒ 40 + 32 + 2

⇒ 74g/mol

Thus, the molar mass of Ca(OH)₂is 74g/mol.

v. Al₂(SO₄)₃

2 × Molar mass of aluminium + 3 × Molar mass of sulphur + 12×molar mass of oxygen

⇒ 2 × 27 + 3 × 32 + 12 × 16

⇒ 54 + 96 + 192

⇒ 342g/mol

Thus, the molar mass of Al₂(SO₄)is 342g/mol.

Mass of oxygen atoms = number of molecules present 1g of hydrogen gas

Number of molecules in 1g H₂ = 1/2 mole

⇒ 1/2×6.022 × 10²³atoms

⇒ 3.01 × 10²³

Thus, the mass possessed by certain number of oxygen atoms is 3.01 × 10²³

Given:

Mass of sodium carbonate = 212g

Sodium carbonate is Na₂CO₃

First, we need to calculate the number of moles of sodium

carbonate we have in a 212g sample. To calculate this, we will

find the molar mass of sodium carbonate (Na₂CO₃):

⇒ 2 × Molar mass of sodium + Molar mass of carbon + 3×molar mass of oxygen

⇒ 2 × 23 + 12 + 3 × 16

⇒ 46 + 12 + 48

⇒ 106g/mol

Thus, the molar mass of Na₂CO₃ is 106g/mol.

Now we will calculate total number of moles by applying the

formula given:

⇒

⇒ 2 mol

Now, we know that every mole of Na₂CO₃ have 2 moles of Na⁺ ions. Hence, total moles of Na₂CO₃ is 4 moles

Number of ions present = 6.022 × 10²³ × 4mol = 24.088 × 10²³ions

Dalton’s symbols were not used in chemistry as he used the symbols for elements in a very specific sense. He also meant a definite quantity of that element by depicting symbol.

But a scientist Berzilius suggested that the symbols of elements be made from one or two letters of the element.

Helium and neon exist as independent atoms. Mostly elements of noble gases exist as independent forms.

Yes, atoms exist in aqueous solution.

For example: Common salt (NaCl) dissolves in water (aqueous

solution) and breaks into Na⁺ and Cl^- ions.

i. The symbol of aluminium is Al.

ii. The symbol of cobalt is Co.

iii. The symbol of arsenic is As.

iv. The symbol of radon is Rn.

The relative atomic mass of an element is defined as the average mass of the atom, as compared to 1/12^ththe mass of one-carbon atom.

Relative atomic mass of sulphur is 32 means the average mass of the sulphur atom is found with respect to an atom of carbon-12.

i. Al₂(CO₃)₃ : There are total two ions of Al³⁺ and three ions of CO₃^2- present. Hence total no. of ions = 5

ii. AlBr₃: There are total three ions of Br ^- and one ion of Al³⁺ present. Hence total no. of ions = 4

It is given that an element has a valency of 3.

Let the element is “X”

Valency of sulphide (S^2-) = -2

Formula for sulphide of element X:

Formula: X₂S₃

The name of the compound represented by formula K₂SO₄ is potassium sulphate.

The formulae of the compounds formed from the sets of carbon and oxygen are:

i. Na₂CO₃ – Sodium carbonate

ii. H₂CO₃ – Carbonic acid

iii. NaHCO₃ – Sodium bicarbonate

It is given that sulphate of a metal is MSO₄.

By applying criss cross method, we can know the valency of “M”:

Hence, the valency of “M” is +2

Now, formula for chloride of metal M:

Formula: MCl₂

Thus, the formula for the chloride of metal is MCl₂.

It is given that carbonate of a metal M is M₂CO₃.

By applying criss cross method, we can know the valency of “M”:

Hence, the valency of “M” is +1

Now, formula for phosphate of metal M:

Formula: M₃PO₄

Thus, the formula for phophate of metal is M₃PO₄.

i. In 1961, for a universally accepted atomic mass unit, carbon-12 isotope was chosen as the standard reference for measuring atomic masses.

ii. One atomic mass unit is a mass unit equal to exactly 1/12^th the mass of one atom of carbon-12.

While searching for various atomic mass units, scientists initially took 1/16 of the mass of a naturally occurring oxygen atom as one unit. The two reasons are:

i. According to them, oxygen reacted with a large number of elements and formed compounds.

ii. The given atomic mass unit gave masses of most of the elements as whole numbers.

The difference between 3O and O₃:

i. 3O indicates the three oxide ions.

ii. O₃ indicates one ozone molecule which is formed by combining of three atoms of oxygen.

iii. O₃ is known as ozone gas which protects the earth from ultraviolet radiations.

It is given that element X has a valency of 3.

Formulafor chloride of element X:

Formula: MCl₃

Thus, the formula for chloride of element X is MCl₃.

Formulafor sulphide of element X:

Formula: M₂S₃

Thus, the formula for sulphide of element X is M₂S₃.

i. Given:

Valency of carbon is 4

Valency of chlorine is 1

By applying criss-cross method:

Formula: CCl₄

Thus, the formula for carbon tetrachloride is CCl₄.

ii. Given:

Valency of element X is 1

Valency of oxygen is 2

By applying criss-cross method:

Formula: X₂O

Let the amount of barium sulphate is “x”g

The given reaction is:

According to law of conservation of mass,

Amount of reactants = Amount of products formed

Hence, 20.8g + 14.2g = x + 11.7g

⇒ 35g = x + 11.7g

⇒ x = 23.5g

Thus, the amount of barium sulphate is 23.5g.

The law involved is Law of conservation of mass which states that

the sum of the masses of reactants and products remain unchanged.

Two monovalent cations: Na⁺ and K⁺

Compounds they make: NaCl and KCl

Two divalent cations: Ca²⁺ and Ba²⁺

Compounds they make: CaCl₂ and BaCl₂

Two trivalent cations: Al³⁺ and Fe³⁺

Compounds they make: AlCl₃ and FeCl₃

Note: Monovalent cations: The cations (positively charged species) which have a valency of +1 are called monovalent cations.

Divalent cations: The cations which have a valency of +2 are called divalent cations.

Trivalent cations: The cations which have a valency of +3 are called trivalent cations.

(i) Sodium carbonate

By applying criss-cross method:

Formula: Na₂CO₃

Thus, the formula for sodium carbonate is Na₂CO₃.

(ii) Magnesium chloride

By applying criss-cross method:

Formula: MgCl₂

Thus, the formula for magnesium chloride is MgCl₂.

(iii) Hydrogen sulphide

By applying criss-cross method:

Formula: H₂S

Thus, the formula for hydrogen sulphide is H₂S.

(i) For Na⁺

By applying criss-cross method:

Formula: NaNO₃

(ii) For K⁺

By applying criss-cross method:

Formula: KNO₃

(iii) For Al³⁺

By applying criss-cross method:

Formula: Al(NO₃)₃

(iv) For Mg²⁺

By applying criss-cross method:

Formula: Mg(NO₃)₂

(v) For Ca²⁺

By applying criss-cross method:

Formula: Ca(NO₃)₂

(vi) For Zn²⁺

By applying criss-cross method:

Formula: Zn(NO₃)₂

(i) Cr³⁺ and SO₄^2-

By applying criss-cross method:

Chemical formula: Cr₂(SO₄)₃

Name of the compound: Chromium sulphate

(ii) Pb²⁺ and NO₃^–

By applying criss-cross method:

Chemical formula: Pb(NO₃)

Name of the compound: Lead nitrate

(iii) Mg²⁺ and CO₃^2–

By applying criss-cross method:

Chemical formula: MgCO₃

Name of the compound: Magnesium carbonate

(i) The two ions are Ca²⁺ (cation) and SO₄^2-(anion)

(ii) When positive ion is replaced by sodium ion:

Chemical formula: Na₂SO₄

(iii) The name of the resulting compound (Na₂SO₄) is sodium sulphate.

(i) Formula unit mass of Na₂CO₃:

2 × (Atomic mass of Na) + atomic mass of C + 3 × (Atomic

mass of oxygen)

⇒ 2 × 23 + 12 + 3×16

⇒ 46 + 12 + 48

⇒ 106u

(ii) To calculate the mass of 1 mole of oxygen atoms:

By applying the formula given:

Mass of oxygen atoms = Number of moles × Molar mass

⇒ Mass of oxygen atoms = 1mole × 32g/mol

⇒ Mass of oxygen atoms = 32g

Thus, the mass of 1 mole of oxygen atoms is 32g.

(iii) Given: Mass of oxygen = 12g

Atomic mass = 16u

By applying the formula given:

⇒

⇒ Number of moles = 0.75mol

Thus, the number of moles = 0.75mol

Chemical formula: X₃P

(ii)

Chemical formula: XCl

(iii) The element X is metal.

Given:

No. of moles = 10mol

Mass of water in each bottle = 20g

As we know that molar mass of water (H₂O) is 18g/mol

Hence, we can say that 1mole of water = 18g

10 moles of water = 18 × 10 = 180g

Number of bottles = 180g /20g = 9

Thus, we will buy 9 bottles.

The two important laws of chemical combination are:

i) Law of constant proportion:

The law states that “in a chemical substance, the elements are always present in definite proportions by mass.

For example: In ammonia (NH₃), nitrogen and hydrogen are always present in the ratio of 14:3 by mass, whatever the method

or source from which it is obtained.

In water, hydrogen and oxygen are always present in the ratio of 1:8 by mass, whatever the method or source from which water is obtained.

ii) Law of conservation of mass:

According to the law of conservation of mass states, atoms are neither created nor destroyed in a chemical reaction. This means that the total mass of the products formed in a chemical reaction must be equal to the mass of reactants consumed.

For example:

Sum of mass of reactants = Sum of masses of prodcuts

(i) Relative atomic mass: The relative atomic mass of an element is defined as the average mass of the atom, as compared to 1/12^ththe mass of one-carbon atom.

The relative atomic masses of all the elements have been found with respect to an atom of carbon-12.

(ii) Atomic mass unit: One atomic mass unit is a mass unit equal to the exactly 1/12^th the mass of one atom of carbon-12.

Carbon-112 was chosen as the standard reference for measuring atomic masses.

(iii) Ions: Compounds composed of metals and non-metals has charged species. The charge species are known as ions.

The negatively charged ions are called anions and positively charged ions are called cations.

(iv) Ionic compounds: Compounds formed by the exchange of ions (cations or anions) are called ionic compounds.

In ionic compounds, the charge on each ion is used to determine the chemical formula of the compound.

(v) Atomicity:

Atomicity: The number of atoms constituting a molecule is known as its atomicity. For example:

In P₄, there are four atoms of phosphorus constituting a molecule. Hence, the atomicity is 4 (tetra-atomic).

(i) Carbon and hydrogen:

CH₄ (methane)

C₂H₆ (ethane)

(ii) Nitrogen and magnesium:

Magnesium nitrate

Formula: Mg₃(NO₃)₂

(iii) Sodium and phosphorus:

Sodium phosphate

Formula: Na₃PO₄

(iv) Potassium and oxygen:

Potassium oxide

Formula: K₂O

(v) Boron and oxygen:

Boron oxide

Formula: B₂O₃

In first experiment:

Mass of barium sulphate = 1.288g

Mass of barium left = 1.03g

Mass of Sulphur present = 1.288 – 1.03 = 0.25g

Percentage of sulphur present in

⇒ Percenatage of sulphur is 20%

In second experiment:

Mass of barium sulphate = 3.672g

Mass of barium left = 2.938g

Mass of Sulphur present = 3.672 – 2.938 = 0.73g

Percentage of sulphur present

⇒ Percenatage of sulphur is 20%

In both the experiments, the percentage of sulphur is same.

Hence, it is verified that these figures verify the law of constant proportions.

Note: Law of constant proportions states that “in a chemical

substance, the elements are always present in definite

proportions by mass.

Given: Number of moles = 5

Atomic mass of AlBO₃ = 30+ 10.8 + 2 × 16 = 72.8g/mol

To find out the mass of aluminium borate, apply the formula given:

⇒

⇒ Mass = 5 mol × 72.8g/mol

⇒ Mass = 364g

Thus, the mass of aluminium borate is 364g.

Given: Number of moles = 2

Atomic mass of KBr = 39 + 80 = 119g/mol

To find out the mass of potassium bromide, apply the formula given:

⇒

⇒ Mass = 2mol × 119g/mol

⇒ Mass = 238g

Thus, the mass of potassium bromide is 238g.

Given: Number of moles = 10

Atomic mass of HCN = 1+ 12 + 14 = 27g/mol

To find out the mass of hydrogen cyanide, apply the formula given:

⇒

⇒ Mass = 10 mol × 27g/mol

⇒ Mass = 270g

Thus, the mass of hydrogen cyanide is 270g.

Given: Number of moles = 2

Atomic mass of CH₄ = 12+ 1 × 4 = 16g/mol

To find out the mass of methane, apply the formula given:

⇒

⇒ Mass = 2 mol × 16g/mol

⇒ Mass = 32g

Thus, the mass of methane is 32g.

Given: Number of moles = 7

Atomic mass of H₂SO₄ = 2 ×1 + 32 + 4 × 16 = 98g/mol

To find out the mass of sulphuric acid, apply the formula given:

⇒

⇒ Mass = 7 mol × 98g/mol

⇒ Mass = 686g

Thus, the mass of sulphuric acid is 686g.

Carbon dioxide

Carbon dioxide is a greenhouse gas and is used to

extinguish fire.

The molar mass of CO₂ is:

Atomic mass of C + 2 × (Atomic mass of oxygen)

⇒ 12 + 2×16

⇒ 44g/mol

The elements present in carbon dioxide gas are carbon and oxygen. The valency of carbon is 1 and the valency of oxygen is 2-

Given:

Mass of gas = 360g

Molar mass = 44g/mol

To calculate the number of moles, we apply the formula:

⇒

⇒ Number of moles = 8.1mol

Thus, the number of moles in 360g of the sample is 8.1

Given: Mass of CO₂ = 88g

To calculate the number of molecules, first we apply the formula:

Number of gram molecules

⇒ Number of gram molecules

⇒ Number of gram molecules = 2

Now, number of molecules = 2 × Avogadro’s number

= 2 × 6.022 × 10²³

= 1.2 × 10²⁴

Thus, number of molecules of CO₂ is 1.2 × 10²⁴

(i) The ratio of gold and copper in the jewellery is 90:10

(ii) Mass of gold in 1g jewellery = 90% × 1g

= 0.9g

Atomic mass of gold = 197g/mol

197 g = 6.022 × 10²³ atoms

⇒ 2.72 × 10²¹ atoms

Thus, the number of atoms present in 1g of gold is 2.72 × 10²¹ atoms.

(iii) Jyoti is aware, concern and a knowledgeable person.

(i) A fruit which contains iron is pomegranate. It is iron rich fruit.

It also contains other important nutrients too.

A vegetable which contains iron is spinach. It is iron rich

vegetable. It contains 3.6mg of iron

Iron is a metal.

(ii) Iron sulphate:

Formula: Fe₂(SO₄)₂ or FeSO₄

(iii) Iron is present in haemoglobin (in blood) which acts as oxygen carrier in our body. It is an important nutrient for our body

Deficiency of iron leads to may problems. It can cause anaemia, shortness of breath, tiredness, headache or migraine. Hence, it must be a part of our diet.

(i) Number of moles is equal to the mass of an element divided by its molar mass. The number of particles (atoms, molecules, ions) present in 1 mole of substance is fixed with a value of 6.022 × 10²³ atoms (Avogadro’s number)

(ii) The unit of Avogadro’s number depends upon the nature the nature of the substance. The unit of Avogadro’s number can be atoms. molecules or ions.

(iii) Mr. Seema is concern for students. She revises all the topics with students so that they can grab the topics easily. She is a very good teacher.

Acoording to the law of conservation of mass, the mass of products formed are equal to the mass of reactants consumed. Hence, the mass of salt and water formed is equal to the combined mass of magnesium hydroxide Mg (OH)₂ and HCl.

In order to verify the law of conservation of mass, we carry out chemical reactions in a closed container, so that gaseous products do not escape. According to the law, mass of products must be equal to the mass of reactants.

The correct chemical formula is BiPO₄

Formula: BiPO₄

The correct chemical formula of (i) is CaCl₂

The correct chemical formula of (iii) is Na₂SO₄

The correct chemical formula of (iv) is Na₂S

Mass of barium sulphate =107.2 g

Mass of barium chloride = 116.1g

Mass of beaker = 150 g

Total mass of reaction mixture in the beaker including mass of

Beaker = 107.2g + 116.1g + 150g = 373.3g

Let the mass of sodium etanoate is “x”g

Acoording to law of conservation of mass,

Sum of masses of reactants = Sum of masses of products

Hence, 10.6g +12.g = x + 4.4g +1.8g

⇒ x = 16.4g

Thus, the mass of sodium ethanoate formed is 16.4g

In experiment (i), the ratio is 2:16, i.e., 1:8

In experiment (ii), the ratio is 4:32, i.e., 1:8

According to the law of constant proportions, the ratio of third experiment should be the same (1:8)

Therefore, the mass of O₂ used during 3^rd experiment is 8g.

If N₂ = 5.6g, H₂ = 1.2g

According to the law of conservation of mas,

5.6g + 1.2g = 6.8g of ammonia

Hence, the option (b) is correct

⇒

⇒ Number of moles = 20

Number of molecules = 20 × 6.022 × 10²³

= 1.2044 × 10²⁵ molecules of water

Thus, option (d) is correct.

Molar mass of sucrose = C₁₂H₂₂O₁₁

= 12× 12 + 1× 22 + 16 × 11 = 342g/mol

⇒

⇒ Number of moles = 0.01

Sucrose (C₁₂H₂₂O₁₁) contains 11 oxygen atoms

⇒ 11 × 6.022 × 10²³

For 0.01 moles of sucrose

⇒ 0.01 × 11 × 6.022 × 10²³= 6.6 × 10²²

Now, Molar mass of water = H₂O = 2× 1 + 16 = 18g/mol

⇒

⇒ Number of moles = 1

Sucrose (H₂O) contains 1 oxygen atom = 6.022 × 10²³

For 1 mole of water = 6.022 × 10²³

Now, add the both values: 6.6 × 10²²+ 6.022 × 10²³

We get 6.68 × 10²³ atoms. Hence, the option (a) is correct.

Less is the molar mass, more will be the number of

molecules. H₂ has lowest molar mass among the given options.

Hence, it contains he maximum number of molecules.