
Chlorine exists in two isotopic forms Cl-37 and Cl-35, but its atomic mass is 35.5. This indicates the ratio of Cl-37 and Cl-35 approximately:
(a)-1: 2
(b)-1: 1
(c)-1: 3
(d)-3: 1
Answer
568.2k+ views
Hint: You can use this formula to get your final answer-
$\text{Average atomic mass = }\sum\limits_{i=1}^{n}{\text{(atomic mass of an isotope) x (fractional abundance)}}$
Solve it carefully the question is asking you to indicate the ratio of Cl-37 and Cl-35, not the vice versa.
Complete Solution :
Isotopes are those whose atomic number is same but the atomic masses are different like the isotopes of hydrogen, i.e., H-1, H-2, H-3. So, there are two isotopes of chlorine atom in which there are two masses 35 and 37, denoted as Cl-35 and Cl-37.
The average atomic masses are calculated by multiplying atomic masses with their respective relative abundance.
To calculate the ratio of abundances of the given isotopes, we use the equation given by average atomic mass, which is:
$\text{Average atomic mass = }\sum\limits_{i=1}^{n}{\text{(atomic mass of an isotope) x (fractional abundance)}}$
Let the fractional abundance for Cl-35 isotope be 'x'.
So, we can say fractional abundance for Cl-37 isotope will be (1 - x)
For Cl-35:
Mass of Cl-35 isotope = 35 u
Fractional abundance of Cl-35 isotope = x
For Cl-37:
Mass of Cl-37 isotope = 37 u
Fractional abundance of Cl-37 isotope = (1 - x)
The average atomic mass of chlorine = 35.5
Now by inserting all values in the above equation, we get:
$35.5=[(35x)+(37(1-x))]$
$35.5=-2x+37$
$x=0.75$
Fractional abundance for Cl-35 isotope = 0.75
Fractional abundance for Cl-37 isotope = 1 - 0.75 = 0.25
- Now just multiple these with 100 to get their percentages,
Percentage abundance for Cl-35 isotope = 75 %
Percentage abundance for Cl-37 isotope = 25 %
The ratio of Cl-37 and Cl-35 isotopes will be = 1: 3
So, the correct answer is “Option C”.
Note: Like isotopes whose atomic number is same but the atomic masses are different, we have isobars in which the different elements have different atomic numbers and have same atomic masses.
$\text{Average atomic mass = }\sum\limits_{i=1}^{n}{\text{(atomic mass of an isotope) x (fractional abundance)}}$
Solve it carefully the question is asking you to indicate the ratio of Cl-37 and Cl-35, not the vice versa.
Complete Solution :
Isotopes are those whose atomic number is same but the atomic masses are different like the isotopes of hydrogen, i.e., H-1, H-2, H-3. So, there are two isotopes of chlorine atom in which there are two masses 35 and 37, denoted as Cl-35 and Cl-37.
The average atomic masses are calculated by multiplying atomic masses with their respective relative abundance.
To calculate the ratio of abundances of the given isotopes, we use the equation given by average atomic mass, which is:
$\text{Average atomic mass = }\sum\limits_{i=1}^{n}{\text{(atomic mass of an isotope) x (fractional abundance)}}$
Let the fractional abundance for Cl-35 isotope be 'x'.
So, we can say fractional abundance for Cl-37 isotope will be (1 - x)
For Cl-35:
Mass of Cl-35 isotope = 35 u
Fractional abundance of Cl-35 isotope = x
For Cl-37:
Mass of Cl-37 isotope = 37 u
Fractional abundance of Cl-37 isotope = (1 - x)
The average atomic mass of chlorine = 35.5
Now by inserting all values in the above equation, we get:
$35.5=[(35x)+(37(1-x))]$
$35.5=-2x+37$
$x=0.75$
Fractional abundance for Cl-35 isotope = 0.75
Fractional abundance for Cl-37 isotope = 1 - 0.75 = 0.25
- Now just multiple these with 100 to get their percentages,
Percentage abundance for Cl-35 isotope = 75 %
Percentage abundance for Cl-37 isotope = 25 %
The ratio of Cl-37 and Cl-35 isotopes will be = 1: 3
So, the correct answer is “Option C”.
Note: Like isotopes whose atomic number is same but the atomic masses are different, we have isobars in which the different elements have different atomic numbers and have same atomic masses.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

