![SearchIcon](https://vmkt.vedantu.com/vmkt/PROD/png/bdcdbbd8-08a7-4688-98e6-4aa54e5e0800-1733305962725-4102606384256179.png)
A black dot used as a full stop at the end of a sentence has a mass of about one attogram. Assuming that the dot is made up of carbon, calculate the approximate number of carbon atoms present in the dot. $\left( {1{\text{ attogram}} = {{10}^{ - 18}}{\text{g}}} \right)$
Answer
452.7k+ views
Hint:To answer this question, you must recall Avogadro's number. The Avogadro’s constant or Avogadro’s number gives the number of molecules/ atoms/ ions present in one mole of a substance. We shall find the moles of carbon present and then multiply it with Avogadro’s number to get the number of atoms.
Formula used:
$n = \dfrac{w}{M}$
Where, $n$ denotes the number of moles of the given substance
$w$ denotes the given mass of the substance
And, $M$ denotes the molar mass of the given substance.
Complete step by step answer:
The value of Avogadro’s constant is $6.023 \times {10^{23}}$ and it is represented as ${N_A}$. Thus, 1 mole of any substance contains $6.023 \times {10^{23}}$ particles.
In the question, we are given a dot with a mass of one attogram. The dot is given to be made up of carbon atoms. We know that one attogram weighs about ${10^{18}}g$.
First, we find the number of moles of carbon present in the dot. We know that it is given by the formula $n = \dfrac{w}{M}$
Substituting the values for carbon atoms, we get, $n = \dfrac{{{{10}^{ - 18}}}}{{12}}$
The number of atoms in n moles of carbon are given by $N = n \times {N_A}$
Substituting the values, we get, $N = \dfrac{{{{10}^{ - 18}}}}{{12}} \times 6.023 \times {10^{23}}$
$ \Rightarrow N = 5.02 \times {10^4}$
So the number of atoms of carbon present in a dot of mass one attogram is $5.02 \times {10^4}$.
Note:
The value of the Avogadro constant is given in such a way so that the mass of one mole of any given chemical substance (in grams) has the same numerical value (for all practical purposes) as the mass of one molecule of the substance in terms of atomic mass units (amu). One atomic mass unit is $1/12th$ of the mass of one carbon- 12 atom. It is approximately equal to the mass of one proton or one neutron.
Formula used:
$n = \dfrac{w}{M}$
Where, $n$ denotes the number of moles of the given substance
$w$ denotes the given mass of the substance
And, $M$ denotes the molar mass of the given substance.
Complete step by step answer:
The value of Avogadro’s constant is $6.023 \times {10^{23}}$ and it is represented as ${N_A}$. Thus, 1 mole of any substance contains $6.023 \times {10^{23}}$ particles.
In the question, we are given a dot with a mass of one attogram. The dot is given to be made up of carbon atoms. We know that one attogram weighs about ${10^{18}}g$.
First, we find the number of moles of carbon present in the dot. We know that it is given by the formula $n = \dfrac{w}{M}$
Substituting the values for carbon atoms, we get, $n = \dfrac{{{{10}^{ - 18}}}}{{12}}$
The number of atoms in n moles of carbon are given by $N = n \times {N_A}$
Substituting the values, we get, $N = \dfrac{{{{10}^{ - 18}}}}{{12}} \times 6.023 \times {10^{23}}$
$ \Rightarrow N = 5.02 \times {10^4}$
So the number of atoms of carbon present in a dot of mass one attogram is $5.02 \times {10^4}$.
Note:
The value of the Avogadro constant is given in such a way so that the mass of one mole of any given chemical substance (in grams) has the same numerical value (for all practical purposes) as the mass of one molecule of the substance in terms of atomic mass units (amu). One atomic mass unit is $1/12th$ of the mass of one carbon- 12 atom. It is approximately equal to the mass of one proton or one neutron.
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The highest possible oxidation states of Uranium and class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Find the value of x if the mode of the following data class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
10 examples of friction in our daily life
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Prokaryotic Cells and Eukaryotic Cells
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
State and prove Bernoullis theorem class 11 physics CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What organs are located on the left side of your body class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How many valence electrons does nitrogen have class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)