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NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers

NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers

Edited By Ramraj Saini | Updated on Feb 29, 2024 05:51 PM IST

Rational Numbers Class 8 Questions And Answers provided here. These NCERT Solutions are created by expert team at craeers360 keeping the latest syllabus and pattern of CBSE 2023-23 . Rational numbers are those numbers which can be represented in p/q form where q ≠0 and p & q are integers. From this definition, we can conclude that all the integers come under the category of a rational number. In this chapter, you will learn about rational numbers, real numbers, whole numbers, integers, and natural numbers and also study their properties like closure, commutativity, associativity. Students can download NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers pdf using the link given below.

In this particular chapter, there are a total of 24 questions in 2 exercises. Students must complete the NCERT Class 8 Math Syllabus and practice all NCERT Solutions for class 9 maths here.

Rational Numbers Class 8 Questions And Answers PDF Free Download

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Rational Numbers Class 8 Solutions - Important Formulae

  • Additive Identity: (a / b + 0) = (a / b).

  • Multiplicative Identity: (a / b) × 1 = (a / b).

  • Multiplicative Inverse: (a / b) × (b / a) = 1.

  • Additive Inverse: a + (-a) = 0.

  • Closure Property – Addition: a + b is a rational number.

  • Closure Property – Subtraction: a - b is a rational number.

  • Closure Property – Multiplication: a × b is a rational number.

  • Closure Property – Division: Rational numbers are not closed under division.

  • Commutative Property – Addition: a + b = b + a.

  • Commutative Property – Multiplication: a × b = b × a.

  • Associative Property – Addition: (a + b) + c = a + (b + c).

  • Associative Property – Multiplication: (a × b) × c = a × (b × c).

  • Distributive Property: a × (b + c) = (a × b) + (a × c).

Free download NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers for CBSE Exam.

Rational Numbers Class 8 NCERT Solutions (Intext Questions and Exercise)


Addition
Subtraction
Multiplication
Division
Rational Numbers
Yes
Yes
...
No
Integers
...
Yes
...
No
Whole Numbers
...
...
Yes
...
Natural Numbers
...
No
... ...

Answer: It can be seen that rational numbers, integers, whole numbers, natural numbers are not closed under division because of Zero is included in these numbers. Any number divided by zero is not defined.


Addition Subtraction Multiplication Division
Rational Numbers Yes Yes Yes No
Integers Yes Yes Yes No
Whole Numbers Yes No Yes No
Natural Numbers Yes No Yes Yes


Q2 Complete the following table:

Commutative for


Addition Subtraction Multiplication Division
Rational Numbers Yes .. ... ...
Integers ... No ... ...
Whole Numbers ... ... Yes ...
Natural Numbers ... ... ... No

Answer: In rational numbers, a \div b \neq b \div a

also a-b \neq b-a


Addition Subtraction Multiplication Division
Rational Numbers Yes No Yes No
Integers Yes No Yes No
Whole Numbers Yes No Yes No
Natural Numbers Yes No Yes No

Q3 Complete the following table:

Associative for


Addition Subtraction Multiplication Division
Rational Numbers ... ... ... No
Integers ... ... Yes ...
Whole Numbers Yes ... ... ...
Natural Numbers ... No ... ...

Answer: For associative in multiplication:- a \times (b \times c) = (a \times b) \times c


Addition Subtraction Multiplication Division
Rational Numbers Yes No Yes No
Integers Yes No Yes No
Whole Numbers Yes No Yes No
Natural Numbers Yes No Yes No


(i) \left \{ \frac{7}{5}\times \left ( \frac{-3}{12} \right ) \right \}+\left \{ \frac{7}{5}\times \frac{5}{12} \right \} (ii) \left \{ \frac{9}{16}\times \frac{4}{12} \right \}+\left \{ \frac{9}{16}\times \frac{-3}{9} \right \}

Answer: (i) Using distributivity, a(b+c) = ab + ac

\left \{ \frac{7}{5}\times \left ( \frac{-3}{12} \right ) \right \}+\left \{ \frac{7}{5}\times \frac{5}{12} \right \} = \frac{7}{5}\left ( \frac{-3}{12}+\frac{5}{12} \right )\\\\\\=\frac{7}{5}\times\frac{1}{6} = \frac{7}{30}

(ii) Using distributivity of multiplication over addition and subtraction,

\left \{ \frac{9}{16}\times \frac{4}{12} \right \}+\left \{ \frac{9}{16}\times \frac{-3}{9} \right \} = \frac{9}{16}\left ( \frac{4}{12}-\frac{3}{9} \right )\\\\\\=\frac{9}{16}\times0 = 0

Q5 Write the rational number for each point labeled with a letter:

Answer: (i) In this, we can see that 1 is divided into 5 parts each, so when we are moving from zero to the right-hand side, it is easy to observe that

All the numbers should contain 5 in their denominator. Thus, A is equal to \frac{1}{5} , B is equal to \frac{4}{5} , C is equal to \frac{5}{5} = 1 , D is equal to \frac{8}{5} , E is equal to \frac{9}{5}

(ii) Here we see that 1 is divided in 6 parts each. So when we move from zero towards left we observe that

All the numbers should contain 6 in their denominator. Thus, F is equal to \frac{-2}{6} , G is equal to \frac{-5}{6} , H is equal to \frac{-7}{6} , I is equal to \frac{-8}{6} , J is equal to \frac{-11}{6}

NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers - Exercise: 1.1

Q1 (i) Using appropriate properties find. (i)\:\frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}-\frac{3}{5}\times \frac{1}{6}

Answer: By using the commutativity property of numbers, we get,

-\frac{2}{3}\times \frac{3}{5}+\frac{5}{2}-\frac{3}{5}\times\frac{1}{6} = -\frac{2}{3}\times \frac{3}{5}-\frac{3}{5}\times\frac{1}{6}+\frac{5}{2}

(Now we will use distributivity of numbers)

= = \frac{3}{5}\left ( \frac{-2}{3}+\frac{-1}{6} \right )+\frac{5}{2}\\\\\\=\frac{3}{5}\times \frac{-5}{6}+\frac{5}{2}\\\\\\=\frac{-1}{2}+\frac{5}{2}\\\\\\=\frac{-1+5}{2} = \frac{4}{2}\\\\=2

Q1 (ii) Using appropriate properties find. (ii)\frac{2}{5}\times \frac{-3}{7}- \frac{1}{6}\times \frac{3}{2}+\frac{1}{14}\times \frac{2}{5}

Answer: By using commutativity, we get

\frac{2}{5}\times \left ( \frac{-3}{7} \right )-\frac{1}{6}\times \frac{3}{2}+\frac{1}{14}\times \frac{2}{5} = \frac{2}{5}\times \left ( \frac{-3}{7} \right ) +\frac{1}{14}\times \frac{2}{5}-\frac{1}{6}\times \frac{3}{2}

Now by distributivity,

= \frac{2}{5}\left ( \frac{-3}{7}+\frac{1}{14} \right )-\frac{1}{4}\\\\\\ =\frac{2}{5}\times \frac{-5}{14}-\frac{1}{4} = \frac{-1}{7}-\frac{1}{4} = \frac{-4-7}{28} = \frac{-11}{28}

Q2 Write the additive inverse of each of the following:

(i) \frac{2}{8} (ii) \frac{-5}{9} (iii) \frac{-6}{-5} (iv) \frac{2}{-9} (v) \frac{19}{-6}

Answer: (i) The additive inverse of \\ \\ \frac{2}{8} is \frac{-2}{8} because \frac{2}{8}+\frac{-2}{8} = \frac{2-2}{8} = 0

(ii) The additive inverse of \frac{-5}{9} is \frac{5}{9} because \frac{-5}{9}+\frac{5}{9} = \frac{-5+5}{9} = 0

(iii) The additive inverse of \frac{-6}{-5} is \frac{6}{-5} because \frac{-6}{-5}+\frac{6}{-5} = \frac{-6+6}{-5} = 0

(iv) The additive inverse of \frac{2}{-9} is \frac{-2}{-9} because \frac{2}{-9}+\frac{-2}{-9} = \frac{2-2}{-9} = 0

(v) The additive inverse of \frac{19}{-6} is \frac{-19}{-6} because \frac{19}{-6}+\frac{-19}{-6} = \frac{19-19}{-6} =0

Q3 Verify that – (– x) = x for (i) x = \frac{11}{15} (ii) x = \frac{-13}{17}

Answer: (i) We have x = \frac{11}{15}

The additive inverse of x = \frac{11}{15} is -x = \frac{-11}{15}

The same equality \frac{11}{15}+\left ( \frac{-11}{15} \right ) = 0

which implies -\left ( \frac{-11}{15} \right ) = \frac{11}{15} shows that -(-x) = x

(ii) Additive inverse of x = \frac{-13}{17} is -x = \frac{13}{17} (since \frac{-13}{17}+\frac{13}{17} = 0 )

The same quality shows that the additive inverse of \frac{13}{17} is \frac{-13}{17}

i.e., -(-x) = x

Q4 Find the multiplicative inverse of the following. (i) - 13 (ii) \frac{-13}{19} (iii) \frac{1}{5} (iv) \frac{-5}{8}\times \frac{-3}{7} (v) -1\times \frac{-2}{5} (vi) - 1

Answer:

(i) The multiplicative inverse of -13 is \frac{-1}{13} because -13\times \frac{-1}{13} = 1

(ii) The multiplicative inverse of \frac{-13}{19} is \frac{-19}{13} because of i

(iii) The multiplicative inverse of \frac{1}{5} is 5 because \frac{1}{5}\times 5 = 1

(iv) The multiplicative inverse of \frac{-5}{8}\times \frac{-3}{7} is \frac{56}{15} because \frac{15}{56}\times \frac{56}{15} = 1

(v) The multiplicative inverse of -1\times \frac{-2}{5} is \frac{5}{2} because \frac{2}{5}\times \frac{5}{2} = 1

(vi) The multiplicative inverse of -1 is -1 because -1\times -1 = 1

Q5 Name the property under multiplication used in each of the following.

(i) \frac{-4}{5}\times 1= 1\times \frac{-4}{5} = \frac{-4}{5} (ii) -\frac{13}{17}\times \frac{-2}{7} = \frac{-2}{7}\times \frac{-13}{17} (iii) \frac{-19}{29}\times \frac{29}{-19} = 1

Answer: (i) Multiplying any number with 1 we get the same number back.

i.e., a \times 1 = 1 \times a = a

Hence 1 is the multiplicative identity for rational numbers.

(ii) Commutativity property states that a \times b = b \times a

(iii) It is the multipicative inverse identity, i.e., a\times\frac{1}{-a} = 1

Q6 Multiply \frac{6}{13} by the reciprocal of \frac{-7}{16} .
Answer: We know that the reciprocal of \frac{-7}{16} is \frac{16}{-7} .
Now, \frac{6}{13}\times \frac{16}{-7} = \frac{-96}{91}


Q7 Tell what property allows you to compute \frac{1}{3}\times \left ( 6\times \frac{4}{3} \right ) as \left ( \frac{1}{3}\times 6 \right )\times \frac{4}{3}

Answer: By the Associativity property for multiplication, we know that a × (b × c) = (a × b) × c. Thus property used here is associativity.


Q8 Is \frac{8}{9} the multiplicative inverse of -1\frac{1}{8} ? Why or why not?

Answer:

We know that A is the multiplicative inverse of B if B \times A = 1

Applying this in given question, we get,

\frac{-9}{8}\times \frac{8}{9} = -1 Thus \frac{8}{9} is not the multiplicative inverse of \frac{-9}{8} .

Q9 Is 0.3 the multiplicative inverse of 3\frac{1}{3} ? Why or why not?
Answer:

We know that if A is multiplicative inverse of B then B \times A = 1

In this question, \frac{10}{3}\times\frac{3}{10} = 1 (Since 0.3 = \frac{3}{10} )

This 0.3 is the multiplicative inverse of \frac{10}{3} .


Q10 Write.

(i) The rational number that does not have a reciprocal.

(ii) The rational numbers that are equal to their reciprocals.

(iii) The rational number that is equal to its negative.

Answer:

(i) Zero(0). We know that reciprocal of A is \frac{1}{A} . So for 0, its reciprocal is not defined.

(ii) 1 and -1 . (Since \frac{1}{1} = 1 and \frac{1}{-1} = -1 )

(iii) Zero,0. (as -0=0)

Q11 Fill in the blanks.

(i) Zero has ________ reciprocal.

(ii) The numbers ________ and ________ are their own reciprocals.

(iii) The reciprocal of – 5 is ________.

(iv) Reciprocal of \frac{1}{x} , where x ≠ 0 is ________.

(v) The product of two rational numbers is always a _______.

(vi) The reciprocal of a positive rational number is ________.

Answer: (i) Zero has no reciprocal as it's reciprocal is not defined.

(ii) The numbers 1 and -1 are their own reciprocals as \frac{1}{1} = 1 and \frac{1}{-1} = -1

(iii) \frac{1}{-5} . We know that reciprocal of A is \frac{1}{A} .

(iv) Since \frac{1}{\frac{1}{x}} = x .

(v) rational number. We know that if p and q are 2 rational numbers then pq is also a rational number.

(vi) Positive. Since reciprocal of A is \frac{1}{A} , now if A is positive then reciprocal is also positive.

Class 8 maths chapter 1 question answer - Exercise: 1.2

Q1 Represent these numbers on the number line. (i) \frac{7}{4} (ii) \frac{-5}{6}

Answer:(i) To represent \frac{7}{4} on a number line, firstly we will divide 1 in 4 parts and draw it on a line such as 1/4, 2/4, 3/4, ........, 9/4. Then will mark the required number.

1643706258551

(ii) To represent \frac{-5}{6} on the number line, firstly we will divide 1 in 6 parts and draw it on the left side of zero on number line such as -1/6, -2/6, .......,-9/4. Then mark the required number on the number line.

1643706272607

Q2 Represent \frac{-2}{11} , \frac{-5}{11} , \frac{-9}{11} on the number line

Answer: We will divide 1 into 11 parts, then start marking numbers on left side of zero such as -1/11, -2/11, -3/11,.........,-12/11. Mark the required numbers on the drawn number line.

1643706315618




Q3 Write five rational numbers which are smaller than 2.
Answer: The 5 rational numbers smaller than 2 can be any number in the form of p/q where q \not\equiv 0. Hence infinite numbers are possible.

Examples of 5 such numbers are 1, 1/3, 0, -1, -2


Q4 Find ten rational numbers between \frac{-2}{5} and \frac{1}{2}
Answer:

Rational numbers between any 2 numbers can easily find out by taking their means.

i.e., For \\\frac{-2}{5} and \\\frac{1}{2}

Their mean is \left ( \frac{-2}{5}+\frac{1}{2} \right )\div 2 = \frac{1}{20} . Hence 1 rational number between \frac{-2}{5} and \frac{1}{2} is \frac{1}{20} .

Now we will find the mean between \frac{-2}{5} and \frac{1}{20} .

This implies a new required rational number is \left ( \frac{-2}{5}+\frac{1}{20} \right )/2 = \frac{-7}{40} .

Similarly, we will find a mean between \frac{1}{20} and \frac{1}{2}

New required rational number is \left ( \frac{1}{20}+\frac{1}{2} \right ) /2 = \frac{11}{40}

Similarly, we will take means of new numbers generated between \frac{-2}{5} and \frac{1}{2} .


Q5 Find five rational numbers between.

(i) \frac{2}{3} and \frac{4}{5} (ii) \frac{-3}{2} and \frac{5}{3} (iii) \frac{1}{4} and \frac{1}{2}
Answer: (i) For finding rational numbers between 2 numbers one method is to find means between the numbers repeatedly.

Another method is:- For \frac{2}{3} and \frac{4}{5}

\frac{2}{3} can be written as \frac{10}{15} \left ( \frac{2}{3}\times \frac{5}{5} = \frac{10}{15}\right )

and \frac{4}{5} can be written as \frac{12}{15} \left ( \frac{4}{5}\times \frac{3}{3} = \frac{12}{15}\right )

Thus numbers between \frac{10}{15} and \frac{12}{15} are the required numbers.

Now since we require 5 numbers in between, thus we multiply the numerator and denominator both by 4.

It becomes numbers between \frac{40}{60} and \frac{48}{60} .

Thus numbers are \frac{41}{60}, \frac{42}{60}, \frac{43}{60}, \frac{44}{60}, \frac{45}{60} .

(ii) Similarly for \frac{-3}{2} and \frac{5}{3}

Required numbers fall between \frac{-9}{6} and \frac{10}{6} \left \{ \left ( \frac{-3}{2}\times \frac{3}{3} \right )= \frac{-9}{6} \right \}

Thus numbers are \frac{-8}{6}, \frac{-7}{6}, \frac{-6}{6}, \frac{-5}{6}, \frac{-4}{6}

(iii) For \frac{1}{4} and \frac{1}{2}

Required numbers lie between \frac{1}{4} and \frac{2}{4} or we can say between \frac{8}{32} and \frac{16}{32}

Thus numbers are \frac{9}{32}, \frac{10}{32}, \frac{11}{32}, \frac{12}{32}, \frac{13}{32}


Q6 Write five rational numbers greater than –2.
Answer: There exist infinitely many rational numbers (can be expressed in the form of p/q where q \not\equiv 0) greater than -2.

Few such examples are -1, -1/2, 0, 1, 1/3 etc.


Q7 Find ten rational numbers between \frac{3}{5} and \frac{3}{4} .
Answer: Finding rational numbers between \frac{3}{5} and \frac{3}{4} is equivalent to find rational numbers between and \frac{15}{20} , since these numbers are obtained by just making their denominators equal.

Further, it is equivalent to find a rational number between \frac{96}{160} and \frac{120}{160}

(We obtained the above numbers by multiplying and dividing numbers by 8 to create a difference of at least 10 numbers).

Thus required numbers are \frac{97}{160},\frac{98}{160},\frac{99}{160},........,\frac{106}{160}

Alternate:- Rational numbers can also be found by taking the mean of the given numbers and the newly obtained number.


Class 8 Maths chapter 1 NCERT solutions - Topics

  • Properties Of Rational Numbers
  • Representation Of Rational Numbers On The Number Line
  • Rational Numbers Between Rational Numbers

NCERT Solutions for Class 8 Maths: Chapter-Wise

Key Features of Class 8 Maths Chapter 1 NCERT Solutions

Detailed Explanations: The solutions fo maths chapter 1 class 8 provide step-by-step and detailed explanations for each problem, making it easy for students to understand the concepts.

Illustrative Diagrams: Diagrams and illustrations are often included to help visualize mathematical concepts in these ch 1 maths class 8 solutions.

Variety of Problems: A wide range of problems and exercises are covered, allowing students to practice and test their understanding of the chapter using class 8 maths ch 1 question answer pdf which can be downloaded freely using link given above in this article.

NCERT Solutions for Class 8: Subject-Wise

How to Use NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers?

  • Clear your concepts about various types of numbers using the text given in the textbook.
  • Learn the proper usage of those concepts while solving a particular problem.
  • Start with the practice exercises implement the learnt concepts.
  • During the practice, keep using NCERT solutions for Class 8 Maths chapter 1 Rational Numbers to boost your preparation.

Keep working hard and happy learning!

Also Check NCERT Books and NCERT Syllabus here:

Frequently Asked Questions (FAQs)

1. What are the important topics of chapter Rational Numbers ?

 Properties of rational numbers like commutativity and associativity, negative of a number, reciprocal, and distributivity of multiplication over addition for rational numbers are the important topics of rational numbers class 8 solutions.


2. How many chapters are there in the CBSE class 8 maths ?

There are 16 chapters starting from rational number to playing with numbers in the CBSE class 8 maths. Studnets can practice rational numbers class 8 NCERT solutions above in this article.

3. Where can I find the complete solutions of NCERT for class 8 ?

Here you will get the detailed NCERT solutions for class 8 by clicking on the link.

4. Which is the official website of NCERT ?

NCERT official is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12.

5. How does the NCERT solutions are helpful ?

Here you will get the detailed NCERT solutions for class 8 maths by clicking on the link.

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0.34\; J

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0.16\; J

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2.45×10−3 kg

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 6.45×10−3 kg

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 9.89×10−3 kg

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12.89×10−3 kg

 

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2,000 \; J - 5,000\; J

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200 \, \, J - 500 \, \, J

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2\times 10^{5}J-3\times 10^{5}J

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20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

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K/2\,

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\; K\;

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zero\;

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K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

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33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

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Option 1)

0.02

Option 2)

3.125 × 10-2

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1.25 × 10-2

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2.5 × 10-2

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decrease twice

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increase two fold

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remain unchanged

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be a function of the molecular mass of the substance.

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Molality

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Weight fraction of solute

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Fraction of solute present in water

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Mole fraction.

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twice that in 60 g carbon

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6.023 × 1022

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half that in 8 g He

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558.5 × 6.023 × 1023

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less than 3

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more than 3 but less than 6

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more than 6 but less than 9

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more than 9

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