# Atomic Structure and Chemical Bonding Exe-4E Numericals Answer Chemistry Class-9 ICSE Selina Publishers

Atomic Structure and Chemical Bonding Exe-4E Numericals Answer Chemistry Class-9 ICSE Selina Publishers Solutions Chapter-4. Step By Step ICSE Selina Concise Solutions of Chapter-4 Atomic Structure and Chemical Bonding with All Exercise including MCQs, Very Short Answer Type, Short Answer Type, Long Answer Type, Numerical and Structured/Application Questions Solved . Visit official Website CISCE for detail information about ICSE Board Class-9.

## Atomic Structure and Chemical Bonding Exe-4E Numericals Answer Chemistry Class-9 ICSE Concise Selina Publishers

 Board ICSE Publications Selina Publication Subject Chemistry Class 9th Chapter-4 Atomic Structure and Chemical Bonding Book Name Concise Topics Solution of Exercise – 4E Numericals Answer Type Academic Session 2023-2024

### Exercise – 4E Numericals Answer Type

Atomic Structure and Chemical Bonding Class-9 Chemistry Concise Solutions

Page-82

#### Question 1.

What is the average atomic mass of bromine if it has 49.7% 79Br35 and 50.3% 81Br35?

Given,

49.7% of 7935Br and

50.3% of 8135Br

These isotopes are in the ratio =

These isotopes are in the ratio = 49.7/50.3 = 497/503

Average atomic mass = (79×497)+(81×503)/1000

= 39263+(40743)/1000

= 80006/1000

= 80.006 amu

= 80 amu

∴ Average atomic mass of bromine = 80 amu

#### Question 2.

A sample of an element A contains two isotopes 168A and 188A. If the average atomic mass of the element is 16.2 amu, calculate the percentage of the two isotopes in this sample.

Let,

the percentage of 168A be X and

percentage of 188A be 100-X

Average atomic mass = (16×X)+[18×(100-X)]/100 = 16.2

⇒ 16X + 1800 – 18X = 1620

⇒ -2X + 1800 = 1620

⇒ -2X = 1620 – 1800

⇒ 2X = 180

⇒ X = 180/2 = 90

∴ 100 – X = 100 – 90 = 10

Hence, percentage of the two isotopes in this sample are 90% of 168A and 10% of 188A

—  : End of Atomic Structure and Chemical Bonding Exe-4E Numericals Class-9 ICSE Chemistry Solutions :–

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