Question

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 1

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}$$\text{x }{\displaystyle \frac{0.16}{15}}$$=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$=6.023\text{x1}{\text{0}}^{23}\text{x}{\displaystyle \frac{0.28}{28}}=6.023\text{x1}{\text{0}}^{21}$ 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 $=6.023\text{x1}{\text{0}}^{23}\text{x}{\displaystyle \frac{0.36}{180}}=6.023\text{x1}{\text{0}}^{20}=12.046\text{x1}{\text{0}}^{20}$ 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 $=6.023\text{x1}{\text{0}}^{23}\text{x}{\displaystyle \frac{1.1}{44}}=1.50\text{x1}{\text{0}}^{22}$ 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.