Write the nuclear composition of each isotope and complete the following table: Isotopes Protons Neutrons H 1 1 D 1 2 T 3 1 C 6 12 C 6 14 Cl 17 35 Cl 17 37

The nuclear composition includes number of protons and neutrons. Therefore, the table can be completed as, Isotopes Protons Neutrons H 1 1 1 0 D 1 2 1 1 T 3 1 3 2 C 6 12 6 6 C 6 14 6 8 Cl 17 35 17 18 Cl 17 37 17 20

The relative atomic mass of boron is 10.8. Calculate the percentage of isotopes B 5 10 and B 5 11 occurring in nature.

The relative atomic mass of boron is 10 . 8 . The percentage of isotopes B 5 10 and B 5 11 occurring in nature can be calculated as, Given, The relative mass of boron = 10 . 8 u. Let us assume that 'y' be the percentage of isotope, B 5 10 in the sample. Therefore, the percentage of isotope B 5 11 in the sample = 100 - y . Average atomic mass = Atomic mass of B 5 10 × its fraction + Atomic mass of B 5 11 × its fraction That is, 10 . 8 = 10 × y 100 + 11 × 100 - y 100 10 . 8 = 10 y 100 + 1100 - 11 y 100 10 . 8 = 10 y 100 + 1100 - 11 y 100 10 . 8 = 10 y + 1100 - 11 y 100 11 y - 10 y = 1100 - 1080 That is, y = 20 % . Therefore, the percentage of B 5 10 = 20 % . Now, the percentage of B 5 11 = 100 - 20 = 80 % .

The relative atomic mass of an element A is found to be 16.2u. There are two isotopes A 8 16 and A 8 18 of the element. Calculate the percentage of these two isotopes present in the element.

The relative atomic mass of an element A is found to be 16 . 2 u. There are two isotopes A 8 16 and A 8 18 of the element. The percentage of these two isotopes present in the element can be calculated as, Given, The relative mass of the element = 16 . 2 u. Let us assume that 'y' be the percentage of isotope, A 8 16 in the sample. Therefore, the percentage of isotope A 8 18 in the sample = 100 - y . Average atomic mass = Atomic mass of A 8 16 × its fraction + Atomic mass of A 8 18 × its fraction That is, 16 . 2 = 16 × y 100 + 18 × 100 - y 100 16 . 2 = 16 y 100 + 1800 - 18 y 100 16 . 2 = 16 y + 1800 - 18 y 100 1620 = 1800 - 2 y 2 y = 1800 - 1620 2 y = 180 y = 90 . That is, y = 90 % . Therefore, the percentage of A 8 16 = 90 % . Now, the percentage of A 8 18 = 100 - 90 = 10 % .

Ordinary chlorine has two isotopes Cl 17 35 and Cl 17 37 in the ratio 3 : 1. Calculate the average atomic mass of chlorine.

(i) Mass contribution of the isotope Cl 17 35 = Atomic mass of Cl 17 35 × its fraction. = 35 u × 75 100 = 26 . 25 u . (ii) Mass contribution of Cl 17 37 = Atomic mass of Cl 17 37 × its fraction. = 35 u × 25 100 = 9 . 25 u . (iii) Atomic mass of Cl = Contribution of isotope Cl 17 35 + Contribution of isotope Cl 17 37 . = 26 . 25 u + 9 . 25 u = 35 . 5 u .

Select the isotopes and isobars from the following: (i) Z 18 40 (ii) Y 17 37 (iii) X 17 35 (iv) K 19 40

Isotopes are the elements having same atomic number but different mass number. Therefore, the isotopes are X 17 35 and Y 17 37 . Isobars are the elements having same mass number but different atomic numbers. Therefore, the isobars are Z 18 40 and K 19 40 .

E M B I B E

Secondary Chemistry>Chapter 4 - Structure of Atom>Problems for Proficiency

R R Misra Solutions for Chapter: Structure of Atom, Exercise 3: Problems for Proficiency

Author:R R Misra

R R Misra Chemistry Solutions for Exercise - R R Misra Solutions for Chapter: Structure of Atom, Exercise 3: Problems for Proficiency

Attempt the free practice questions on Chapter 4: Structure of Atom, Exercise 3: Problems for Proficiency with hints and solutions to strengthen your understanding. Secondary Chemistry solutions are prepared by Experienced Embibe Experts.

Questions from R R Misra Solutions for Chapter: Structure of Atom, Exercise 3: Problems for Proficiency with Hints & Solutions

EASY

9th CBSE

IMPORTANT

Secondary Chemistry>Chapter 4 - Structure of Atom>Problems for Proficiency>Q 1

Define the isotopes of an element.

EASY

9th CBSE

IMPORTANT

Secondary Chemistry>Chapter 4 - Structure of Atom>Problems for Proficiency>Q 2

Chemical properties of all the isotopes of an element are similar. Give one reason.

HARD

9th CBSE

IMPORTANT

Secondary Chemistry>Chapter 4 - Structure of Atom>Problems for Proficiency>Q 3

Compositions of the nuclei of two atoms A and B are given as:

(a) A has 6 protons, 6 neutrons
(b) B has 6 protons, 8 neutrons.
(i) What are the mass numbers of A and B?
(ii) What is the relation between A and B?
(iii) Which element or elements do these represent?

HARD

9th CBSE

IMPORTANT

Secondary Chemistry>Chapter 4 - Structure of Atom>Problems for Proficiency>Q 4

Write the nuclear composition of each isotope and complete the following table:

Isotopes	Protons	Neutrons
${}_{1}^{1}H$
${}_{1}^{2}D$
${}_{3}^{1}T$
${}_{6}^{12}C$
${}_{6}^{14}C$
${}_{17}^{35}{Cl}$
${}_{17}^{37}{Cl}$

HARD

9th CBSE

IMPORTANT

Secondary Chemistry>Chapter 4 - Structure of Atom>Problems for Proficiency>Q 5

The relative atomic mass of boron is 10.8. Calculate the percentage of isotopes ${}_{5}^{10}B$ and ${}_{5}^{11}B$ occurring in nature.

HARD

9th CBSE

IMPORTANT

Secondary Chemistry>Chapter 4 - Structure of Atom>Problems for Proficiency>Q 6

The relative atomic mass of an element A is found to be 16.2u. There are two isotopes ${}_{8}^{16}A$ and ${}_{8}^{18}A$ of the element. Calculate the percentage of these two isotopes present in the element.

HARD

9th CBSE

IMPORTANT

Secondary Chemistry>Chapter 4 - Structure of Atom>Problems for Proficiency>Q 7

Ordinary chlorine has two isotopes ${}_{17}^{35}{Cl}$ and ${}_{17}^{37}{Cl}$ in the ratio 3 : 1. Calculate the average atomic mass of chlorine.

HARD

9th CBSE

IMPORTANT

Secondary Chemistry>Chapter 4 - Structure of Atom>Problems for Proficiency>Q 8

Select the isotopes and isobars from the following:

(i) ${}_{18}^{40}Z$

(ii) ${}_{17}^{37}Y$