Answer
403.8k+ views
Hint: To find the solution of the first question, as the required data is already given, you need to only apply the direct formula for the average or relative atomic mass and for the second question, remember mass of one-gram atom is equal to the mass of one mole of atoms of an element.
Complete step by step answer:
(a) Given that,
The relative abundance of silicon with atomic number $28$ is $92.25%$.
The relative abundance of silicon with atomic number $29$ is $4.65%$.
The relative abundance of silicon with atomic number $30$ is $3.10%$.
We should know that, while doing any average mass calculations for an element, it should be equal to the sum of the masses of each isotope and each multiplied by its natural relative abundances.
So, here the atomic number of each isotope of silicon will be equal to its atomic mass.
Thus, the average atomic mass of an element having isotope equals the sum of mass of each isotope multiplied with abundance. So, the formula will be like,
$Average\text{ atomic mass}=\dfrac{\text{(abundance}\times \text{atomic mass) of 1st isotope +}...\text{ }}{100}$
So, Average atomic mass of silicon $=\dfrac{28\times 92.25+29\times 4.65+30\times 3.10}{100}=\dfrac{2810.85}{100}=28.10$
Hence, the average atomic mass of silicon is $28.10g$.
(b) As per the given question, we have to find out the mass of $1.6g$ atoms of oxygen atoms.
We know that, gram atomic mass refers to the mass of one mole of an element in grams, which is its molar mass.
So, the mass of $1g$ atom is equal to one mole of the element which is equal to the molar mass of that element.
Therefore, the mass of $1g$ atom of oxygen will be equal to $16g$.
Thus, the mass of $1.6g$ atom of oxygen will be equal to $(1.6\times 16)g=25.6g$
Hence, the mass of $1.6g$ atom of oxygen is $25.6g$.
Note: The average atomic mass for an element is calculated by summing the masses of the element’s isotopes, each multiplied by its natural abundance on earth. The mass of a one-gram atom equals the mass of one mole of elements in grams, which is its molar mass.
Complete step by step answer:
(a) Given that,
The relative abundance of silicon with atomic number $28$ is $92.25%$.
The relative abundance of silicon with atomic number $29$ is $4.65%$.
The relative abundance of silicon with atomic number $30$ is $3.10%$.
We should know that, while doing any average mass calculations for an element, it should be equal to the sum of the masses of each isotope and each multiplied by its natural relative abundances.
So, here the atomic number of each isotope of silicon will be equal to its atomic mass.
Thus, the average atomic mass of an element having isotope equals the sum of mass of each isotope multiplied with abundance. So, the formula will be like,
$Average\text{ atomic mass}=\dfrac{\text{(abundance}\times \text{atomic mass) of 1st isotope +}...\text{ }}{100}$
So, Average atomic mass of silicon $=\dfrac{28\times 92.25+29\times 4.65+30\times 3.10}{100}=\dfrac{2810.85}{100}=28.10$
Hence, the average atomic mass of silicon is $28.10g$.
(b) As per the given question, we have to find out the mass of $1.6g$ atoms of oxygen atoms.
We know that, gram atomic mass refers to the mass of one mole of an element in grams, which is its molar mass.
So, the mass of $1g$ atom is equal to one mole of the element which is equal to the molar mass of that element.
Therefore, the mass of $1g$ atom of oxygen will be equal to $16g$.
Thus, the mass of $1.6g$ atom of oxygen will be equal to $(1.6\times 16)g=25.6g$
Hence, the mass of $1.6g$ atom of oxygen is $25.6g$.
Note: The average atomic mass for an element is calculated by summing the masses of the element’s isotopes, each multiplied by its natural abundance on earth. The mass of a one-gram atom equals the mass of one mole of elements in grams, which is its molar mass.
Recently Updated Pages
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x1x2xn be in an AP of x1 + x4 + x9 + x11 + x20-class-11-maths-CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the type of food and mode of feeding of the class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples of Material nouns Abstract nouns Common class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)