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?

Answer:

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.

Answer:

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