Worked Solutions - EDCO



Worked Solutions

Chapter 2

Question 28

State the number of protons, electrons and neutrons for each of the

11 28 7 40 19

following: (a) B (b) Si (c) Li (d) Ar (e) F.

5 14 3 18 9

Answer:

a) The atomic number Z = 5, so there are 5 protons and 5 electrons. The mass number A = 11. Number of neutrons = A – Z = 11 - 5 = 6.

b) The atomic number Z = 14, so there are 14 protons and 14 electrons. The mass number A = 28. Number of neutrons = A – Z = 28 - 14 = 14.

c) The atomic number Z =3, so there are 3 protons and 3 electrons. The mass number A = 7. Number of neutrons = A – Z = 7 - 3 = 4.

d) The atomic number Z = 18, so there are 18 protons and 18 electrons. The mass number A = 40. Number of neutrons = A – Z = 40 - 18 = 22.

e) The atomic number Z = 9, so there are 9 protons and 9 electrons. The mass number A = 19. Number of neutrons = A – Z = 19 - 9 = 10.

Question 29

Indicate the mass number and atomic number for atoms which have (a) 19 protons and 20 neutrons (b) 6 protons and 6 neutrons (c) 4 protons and 5 neutrons.

Answer:

(a) Since the number of protons is 19, the atomic number Z = 19. The mass number = number of protons + number of neutrons = 19 + 20 = 39.

(b) Since the number of protons is 6, the atomic number Z = 6. The mass number = number of protons + number of neutrons = 6 + 6 = 12.

(c) Since the number of protons is 4, the atomic number Z = 4. The mass number = number of protons + number of neutrons = 4 + 5 = 9.

Question 37

Calculate the relative atomic mass of silicon with the following isotopic composition: silicon-28 = 92.2%, silicon-29 = 4.7%, silicon-30 = 3.1%.

Answer:

Ar(Si) = [(92.2 x 28) + (4.7 x 29) + (3.1 x 30)] / 100

= [2581.6 + 136.3 + 93] / 100

= 2810.9 / 100

=28.11

Question 38

Calculate the relative atomic mass of copper with the following isotopic composition: copper-63 = 69%, copper-65 = 31%.

Answer:

Ar(Cu) = [(69 x 63) + (31 x 65)] / 100

= [4347 + 2015] / 100

= 6362 / 100

= 63.62

Question 44

Naturally occurring silver contains isotopes of mass numbers 107 and 109, with percentage abundances of 51 and 49 respectively. Calculate the relative atomic mass of silver.

Answer:

Ar(Ag) = [(51 x 107) + (49 x 109)] / 100

= [5457 + 5341] / 100

= 10798 / 100

= 107.98

Question 45

A solid element, X, was found to consist of 48% 151X and 52% 153X. Calculate the relative atomic mass of X and identify the element.

Answer:

Ar(X) = [(48 x 151) + (52 x 153)] / 100

= [7248 + 7956] / 100

= 15204 / 100

= 152.04

The element of relative atomic mass 152 is europium, Eu.

Question 46

Lead consists of 1.5% lead-204, 23.6% lead-206, 22.6% lead-207 and 52.3% lead-208. Calculate the relative atomic mass of lead.

Answer:

Ar(Pb) = [(1.5 x 204) + (23.6 x 206) + (22.6 x 207) + (52.3 x 208] / 100

= [306 + 4861.6 + 4678.2 + 10878.4] / 100

= 20724.2 / 100

=207.24

Question 47

Atom A has a mass number of 36 and an atomic number of 16. Atom B has a mass number of 36 and an atomic number of 18.

a) How many protons has atom A?

(b) How many neutrons has atom B?

(c) Are atoms A and B isotopes of the same element?

Answer:

a) Atom A has 16 protons since its atomic number is 16.

b) Number of neutrons of atom B = A – Z = 36 - 18 = 18

c) A and B are not atoms of the same element since they have different atomic numbers and consequently different numbers of protons.

Question 48

(d) Bromine contains isotopes of mass numbers 79 and 81, with percentage abundances of 55 and 45 respectively. Calculate the relative atomic mass of bromine.

Answer:

Ar(Br) = [(55 x 79) + (45 x 81)] / 100

= [4345 + 3645] / 100

= 7990 / 100

= 79.90

Question 49

(a) State the number of protons, neutrons and electrons in an atom

of the isotope 11B5.

(b) Using a mass spectrometer it was found that boron consisted of 81% boron-11 and 19% boron-10. Calculate the relative atomic mass of boron.

Answer:

a) 11

An atom of B has an atomic number Z = 5, and a mass number A = 11.

5

Z = number of protons = number of electrons = 5.

Number of neutrons = A – Z = 11 – 5 = 6.

(b) Ar(B) = [(81 x 11) + (19 x 10)] / 100

= [891 + 190] / 100

= 1081 / 100

= 10.81

Question 50

Natural chromium in chromium ores contains 4% chromium-50, 84% chromium-52, 10% chromium 53 and 2% chromium-54.

a) What is the atomic number of chromium?

b) What are the mass numbers of the isotopes of chromium?

c) Write down the number of protons, neutrons and electrons in each of the chromium isotopes.

d) Calculate the relative atomic mass of chromium.

Answer:

a) From the periodic table, the atomic number of chromium is 24.

b) The mass numbers of the isotopes of chromium are 50, 52, 53 and 54 respectively.

c) Chromium 50:

The atomic number Z = 24, so there are 24 protons and 24 electrons.

The mass number A = 50.

Number of neutrons = A – Z = 50 – 24 = 26.

Chromium 52:

The atomic number Z = 24, so there are 24 protons and 24 electrons.

The mass number A = 52.

Number of neutrons = A – Z = 52 – 24 = 28

Chromium 53:

The atomic number Z = 24, so there are 24 protons and 24 electrons.

The mass number A = 53

Number of neutrons = A – Z = 53 – 24 = 29.

Chromium 54:

The atomic number Z = 24, so there are 24 protons and 24 electrons.

The mass number A = 54.

Number of neutrons = A – Z = 54 – 24 = 30.

d) Ar(Cr) = [(4 x 50) + (84 x 52) + (10 x 53) + (2 x 54)] / 100

= [200 + 4368 + 530 + 108] / 100

= 5206 / 100

= 52.06

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