
Which of the following contains the maximum number of atoms?
(a). 0.16g $\text{C}{{\text{H}}_{3}}$
(b). 0.28g ${{\text{N}}_{2}}$
(c). 0.36g ${{\text{C}}_{6}}{{\text{H}}_{12}}{{\text{O}}_{6}}$
(d). 1.1g $\text{C}{{\text{O}}_{2}}$
Answer
125.4k+ views
Hint: We can calculate by means of using Avogadro’s number. The number of units in one mole of any substance is called Avogadro’s number or Avogadro’s constant. It is equal to $6.023\text{x1}{{\text{0}}^{23}}$. The units may be electrons, atoms or molecules depending on the nature of substance.
Complete step by step solution:
Molar mass is the mass of one molecule. We can calculate it by adding the atomic mass of each atom in the molecule together. The units for molar mass are g/mol (grams per mole)
We should also know about the mole concept. Mole (mol) is a standard unit that measures the mass of one molecule. The mass of a mol will be different based on the molecule.
This is how the maximum number of atoms is calculated.
0.16g $\text{C}{{\text{H}}_{3}}$
Total molar mass of $\text{C}{{\text{H}}_{3}}$=15g
Since,15gof $\text{C}{{\text{H}}_{3}}$molecules contain$=6.023\text{x1}{{\text{0}}^{23}}$ molecules.
Therefore, 0.16g of$\text{C}{{\text{H}}_{3}}$ contains$=6.023\text{x1}{{\text{0}}^{23}}$$\text{x }\dfrac{0.16}{15}$$=6.4\text{x1}{{\text{0}}^{21}}$molecules.
0.28g ${{\text{N}}_{2}}$
Total molar mass of ${{\text{N}}_{2}}$ = 28g
Since, 28g of${{\text{N}}_{2}}$molecules contain$=6.023\text{x1}{{\text{0}}^{23}}$molecules
Therefore, 0.28g of ${{\text{N}}_{2}}$contains$=6.023\text{x1}{{\text{0}}^{23}}\text{x}\dfrac{0.28}{28}=6.023\text{x1}{{\text{0}}^{21}}$ molecules
0.36g ${{\text{C}}_{6}}{{\text{H}}_{12}}{{\text{O}}_{6}}$
Total molar mass of ${{\text{C}}_{6}}{{\text{H}}_{12}}{{\text{O}}_{6}}$ =180g
Since, 180g of ${{\text{C}}_{6}}{{\text{H}}_{12}}{{\text{O}}_{6}}$ molecules contain$=6.023\text{x1}{{\text{0}}^{23}}$ molecules
Therefore, 0.36g of${{\text{C}}_{6}}{{\text{H}}_{12}}{{\text{O}}_{6}}$ contains $=6.023\text{x1}{{\text{0}}^{23}}\text{x}\dfrac{0.36}{180}=6.023\text{x1}{{\text{0}}^{20}}=12.046\text{x1}{{\text{0}}^{20}}$ molecules.
1.1g $\text{C}{{\text{O}}_{2}}$
Total molar mass of $\text{C}{{\text{O}}_{2}}$=44g
Since, 44f of $\text{C}{{\text{O}}_{2}}$ molecules contain$\text{C}{{\text{O}}_{2}}$molecules
Therefore, 1.1g of $\text{C}{{\text{O}}_{2}}$ molecules contain $ = 6.023\text{x1}{{\text{0}}^{23}}\text{x}\dfrac{1.1}{44}=1.50\text{x1}{{\text{0}}^{22}}$ molecules
So, as per the above calculation, maximum number of molecules from the above species is option (c)
1.1g $\text{C}{{\text{O}}_{2}}$=$1.5\text{x1}{{\text{0}}^{22}}$ molecules.
Note: We should also know about the mole concept. Mole (mol) is a standard unit that measures the mass of one molecule. The mass of a mol will be different based on the molecule.
Complete step by step solution:
Molar mass is the mass of one molecule. We can calculate it by adding the atomic mass of each atom in the molecule together. The units for molar mass are g/mol (grams per mole)
We should also know about the mole concept. Mole (mol) is a standard unit that measures the mass of one molecule. The mass of a mol will be different based on the molecule.
This is how the maximum number of atoms is calculated.
0.16g $\text{C}{{\text{H}}_{3}}$
Total molar mass of $\text{C}{{\text{H}}_{3}}$=15g
Since,15gof $\text{C}{{\text{H}}_{3}}$molecules contain$=6.023\text{x1}{{\text{0}}^{23}}$ molecules.
Therefore, 0.16g of$\text{C}{{\text{H}}_{3}}$ contains$=6.023\text{x1}{{\text{0}}^{23}}$$\text{x }\dfrac{0.16}{15}$$=6.4\text{x1}{{\text{0}}^{21}}$molecules.
0.28g ${{\text{N}}_{2}}$
Total molar mass of ${{\text{N}}_{2}}$ = 28g
Since, 28g of${{\text{N}}_{2}}$molecules contain$=6.023\text{x1}{{\text{0}}^{23}}$molecules
Therefore, 0.28g of ${{\text{N}}_{2}}$contains$=6.023\text{x1}{{\text{0}}^{23}}\text{x}\dfrac{0.28}{28}=6.023\text{x1}{{\text{0}}^{21}}$ molecules
0.36g ${{\text{C}}_{6}}{{\text{H}}_{12}}{{\text{O}}_{6}}$
Total molar mass of ${{\text{C}}_{6}}{{\text{H}}_{12}}{{\text{O}}_{6}}$ =180g
Since, 180g of ${{\text{C}}_{6}}{{\text{H}}_{12}}{{\text{O}}_{6}}$ molecules contain$=6.023\text{x1}{{\text{0}}^{23}}$ molecules
Therefore, 0.36g of${{\text{C}}_{6}}{{\text{H}}_{12}}{{\text{O}}_{6}}$ contains $=6.023\text{x1}{{\text{0}}^{23}}\text{x}\dfrac{0.36}{180}=6.023\text{x1}{{\text{0}}^{20}}=12.046\text{x1}{{\text{0}}^{20}}$ molecules.
1.1g $\text{C}{{\text{O}}_{2}}$
Total molar mass of $\text{C}{{\text{O}}_{2}}$=44g
Since, 44f of $\text{C}{{\text{O}}_{2}}$ molecules contain$\text{C}{{\text{O}}_{2}}$molecules
Therefore, 1.1g of $\text{C}{{\text{O}}_{2}}$ molecules contain $ = 6.023\text{x1}{{\text{0}}^{23}}\text{x}\dfrac{1.1}{44}=1.50\text{x1}{{\text{0}}^{22}}$ molecules
So, as per the above calculation, maximum number of molecules from the above species is option (c)
1.1g $\text{C}{{\text{O}}_{2}}$=$1.5\text{x1}{{\text{0}}^{22}}$ molecules.
Note: We should also know about the mole concept. Mole (mol) is a standard unit that measures the mass of one molecule. The mass of a mol will be different based on the molecule.
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