10th Science Unit 7 Atoms and Molecules Book Back Answers

Options a and d are the same: 1 mole of He = $6.023 \times 10^{23}$ atoms = 4 g.

Option c is 2 g of He, which is less than 4 g.

A single atom of He has a mass of only about 4 amu ($6.6 \times 10^{-24}$ g), far smaller than any of the others.

Atomicity is the number of atoms present in a molecule.

$\mathrm{CO_2}$ has 1 carbon and 2 oxygen atoms, so its atomicity is 3 — a heterotriatomic molecule.

Glucose ($\mathrm{C_6H_{12}O_6}$) is polyatomic, helium is monoatomic and hydrogen ($\mathrm{H_2}$) is diatomic.

Molecular mass of $\mathrm{CO_2} = 12 + (16 \times 2) = 44$, so 1 mole = 44 g.

Number of moles $= 4.4/44 = 0.1$ mole.

Volume $=$ number of moles $\times$ molar volume $= 0.1 \times 22.4 = 2.24$ litre.

Gram atomic mass of nitrogen is 14 g, and one mole of atoms of an element is equal to its gram atomic mass.

So 1 mole of nitrogen atoms weighs 14 g (28 g would be 1 mole of $\mathrm{N_2}$ molecules).

Atomic mass unit (amu) is defined as one-twelfth of the mass of a carbon-12 atom.

Carbon-12 is taken as the standard for measuring atomic and molecular masses.

One mole of hydrogen gas ($\mathrm{H_2}$) contains Avogadro's number of molecules, not atoms.

Since hydrogen is diatomic, one mole of $\mathrm{H_2}$ contains $2 \times 6.023 \times 10^{23}$ atoms.

The other three statements are correct, so statement c is incorrect.

One mole of any gas occupies 22.4 litre at S.T.P; this is the molar volume.

The value is the same for monoatomic, diatomic or polyatomic gases.

For $_{20}\mathrm{Ca}^{40}$, atomic number 20 gives the number of protons = 20.

Mass number 40 = protons + neutrons, so neutrons $= 40 - 20 = 20$.

Electrons are not present in the nucleus.

Oxygen exists as a homodiatomic molecule $\mathrm{O_2}$.

Molecular mass $= 16 \times 2 = 32$; expressed in grams it is the gram molecular mass = 32 g.

One mole of a substance contains as many particles as there are atoms in exactly 12 g of carbon-12.

This number is Avogadro's number, $6.023 \times 10^{23}$.

Isobars have the same mass number (same total of protons + neutrons).

But their atomic numbers differ, so they are atoms of different elements, e.g. $_{18}Ar^{40}$ and $_{20}Ca^{40}$.

Modern atomic theory states that an atom is no longer indestructible.

One element can be changed into another element through artificial transmutation.

Protons and neutrons contribute almost all the mass of an atom.

Their total number is called the mass number.

Relative atomic mass (Ar) is measured on the C-12 scale.

It is also called the Standard Atomic Weight.

From the table of atomic masses, hydrogen (atomic number 1) has an average atomic mass of 1.008 amu.

It is not exactly 1 because the natural isotopes of hydrogen are taken into account.

A molecule made of the same kind of atoms is a homoatomic molecule, e.g. $O_2$, $O_3$.

A molecule made of atoms of different elements is heteroatomic, e.g. HCl.

The number of atoms joined together in one molecule is its atomicity.

E.g. $O_2$ has atomicity 2, $O_3$ has atomicity 3.

One mole of any gas at S.T.P. occupies the molar volume, 22.4 litre.

22.4 litre = 22400 ml.

Phosphorus exists as $P_4$ molecules.

So four atoms are present in one molecule, making its atomicity 4 (polyatomic).

Atoms of two elements can combine in different ratios to form different compounds.

Example: carbon and oxygen combine to form carbon monoxide (CO) and carbon dioxide ($\mathrm{CO_2}$).

Atomicity is the number of atoms present in a molecule.

Noble gases exist as single, unbonded atoms, so their atomicity is one.

Relative atomic mass is only a ratio, so it has no unit.

But when that atomic mass is expressed in grams it is called gram atomic mass, so it has the unit gram (e.g. gram atomic mass of oxygen = 16 g).

One mole of any substance contains Avogadro's number of particles, $6.023 \times 10^{23}$.

So 1 mole of gold and 1 mole of silver each contain $6.023 \times 10^{23}$ atoms, though their masses differ.

Molar mass of $\mathrm{CO_2}$ = (1 × 12) + (2 × 16).

= 12 + 32 = 44 g.

Relative atomic mass of an element is the ratio between the average mass of its atom and 1/12th of the mass of a carbon-12 atom; for aluminium this ratio is 27, so the assertion is true.

The reason restates this definition correctly — one aluminium atom is 27 times heavier than 1/12th of a C-12 atom, which is exactly why its relative atomic mass is 27.

Since the reason gives the correct explanation of the assertion, option (a) is chosen.

Chlorine is diatomic ($Cl_2$), so its relative molecular mass is $2 \times 35.5 = 71$. The value 35.5 is the average atomic mass of chlorine, not its molecular mass; also relative mass is only a ratio and has no unit (amu). Hence the assertion is wrong.

The reason is correct: chlorine exists as the isotopes $_{17}Cl^{35}$ (75%) and $_{17}Cl^{37}$ (25%), whose natural abundances are not equal, and this is why the average atomic mass works out to 35.5.

So A is wrong while R is correct — option (c).

Relative atomic mass of an element is the ratio between the average mass of its isotopes to $\frac{1}{12}$th part of the mass of a carbon-12 atom.

$A_r = \dfrac{\text{Average mass of the isotopes of the element}}{\frac{1}{12}\text{th of the mass of one carbon-12 atom}}$

It is denoted by $A_r$ and is also called 'Standard Atomic Weight'.

Being only a ratio, it has no unit.

Oxygen exists in nature as a mixture of three stable isotopes.

$_8O^{16}$ – mass 15.9949 amu – abundance 99.757 %

$_8O^{17}$ – mass 16.9991 amu – abundance 0.038 %

$_8O^{18}$ – mass 17.9992 amu – abundance 0.205 %

Using these values the average atomic mass of oxygen works out to 15.999 amu.

A molecule is formed when two or more atoms are held together by chemical bonds.

The number of atoms present in a molecule is called its atomicity.

Examples: $He$ – monoatomic (1 atom), $O_2$ – diatomic (2 atoms), $O_3$ – triatomic (3 atoms), $P_4$ – polyatomic (more than 3 atoms).

A heteroatomic molecule contains atoms of different elements; if it has two atoms it is heterodiatomic.

Hydrogen chloride ($HCl$) – one hydrogen atom and one chlorine atom, atomicity two.

Carbon monoxide ($CO$) – one carbon atom and one oxygen atom, atomicity two.

One mole of any gas contains $6.023 \times 10^{23}$ molecules.

The volume occupied by one mole of any gas at S.T.P. (273.15 K and 1 atm) is the same for all gases.

This volume is 22.4 litre (22400 ml) and is called the molar volume or gram molar volume.

Given: ammonia $NH_3$; atomic masses $N = 14$, $H = 1$.

Molecular mass of $NH_3 = 14 + (1 \times 3) = 17$ g.

Formula: Mass % of an element $= \dfrac{\text{mass of that element in the compound}}{\text{molecular mass of the compound}} \times 100$

Substitution: % of N $= \dfrac{14}{17} \times 100$

Result: % of nitrogen in ammonia $= 82.35\,\%$

Mass of 1 mole of nitrogen ($\mathrm{N_2}$): Molecular mass = $2 \times 14 = 28$, so 1 mole = 28 g.

Mass of 3 moles of hydrogen ($\mathrm{H_2}$): Molecular mass of $\mathrm{H_2} = 2 \times 1 = 2$ g; 3 moles = $3 \times 2 =$ 6 g.

Mass of 2 moles of ammonia ($\mathrm{NH_3}$): Molecular mass = $14 + (3 \times 1) = 17$ g; 2 moles = $2 \times 17 =$ 34 g.

Completed equation: 1 mole of nitrogen (28 g) + 3 moles of hydrogen (6 g) → 2 moles of ammonia (34 g).

Verification (law of conservation of mass): Mass of reactants $= 28 + 6 = 34$ g = mass of product $= 34$ g.

(i) 27 g of Al — Given mass = 27 g; atomic mass of aluminium = 27.

Number of moles = $\dfrac{\text{Mass of the element}}{\text{Atomic mass}} = \dfrac{27}{27} = 1$ mole of aluminium.

(ii) $1.51 \times 10^{23}$ molecules of $\mathrm{NH_4Cl}$ — Given number of molecules = $1.51 \times 10^{23}$; Avogadro's number = $6.023 \times 10^{23}$.

Number of moles = $\dfrac{\text{Number of molecules}}{\text{Avogadro's number}} = \dfrac{1.51 \times 10^{23}}{6.023 \times 10^{23}}$.

= $\dfrac{1}{4} = 0.25$ mole of $\mathrm{NH_4Cl}$.

Given: number of moles of each molecule. Formula: Mass = Number of moles × Gram molecular mass.

i. $\mathrm{H_2}$: gram molecular mass $= 1 \times 2 = 2$ g. Mass $= 2 \times 2 = 4$ g.

ii. $\mathrm{Cl_2}$: gram molecular mass $= 35.5 \times 2 = 71$ g. Mass $= 3 \times 71 = 213$ g.

iii. $\mathrm{S_8}$: gram molecular mass $= 32 \times 8 = 256$ g. Mass $= 5 \times 256 = 1280$ g.

iv. $\mathrm{P_4}$: gram molecular mass $= 31 \times 4 = 124$ g. Mass $= 4 \times 124 = 496$ g.

Given: $\mathrm{Al_2(SO_4)_3}$; Al = 27, S = 32, O = 16.

Molecular mass $= (27 \times 2) + [32 + (16 \times 4)] \times 3 = 54 + (96 \times 3) = 54 + 288 = 342$ g.

Number of oxygen atoms $= 4 \times 3 = 12$, so mass of oxygen $= 12 \times 16 = 192$ g.

Mass % of O $= \dfrac{192}{342} \times 100$.

$= 56.14\%$.