The Question is from neet-chemistry basic-chemistry-concepts mole-concept, and it was created by the author Mr. S M HUSSAIN

Question

Which of the following contains the largest number of atoms?

Mr. S M HUSSAIN · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
 **Primary Concept**: Mole concept and Avogadro's number
 **Related Topics**: Atomic mass, molar mass
 **Prerequisites**: Understanding of moles, Avogadro's number ($6.022 	imes 10^{23}$), and calculating molar mass.
## SOLUTION PATHWAY
**Given Information**:
  1 g samples of $O_2$, $N_2$, $H_2$, and $Li$.
  Important conditions: We need to determine which sample contains the largest number of atoms.
**Method & Steps**:
 1. Calculate the number of moles for each substance.
  Reasoning: The number of moles is given by the formula:  $n = \frac{mass}{molar\,mass}$ 
  Formula/Rule: $n = \frac{m}{M}$ where n is the number of moles, m is the mass, and M is the molar mass.
 2. Calculate the number of atoms for each substance.
  Working:  The number of atoms is given by: $Number\,of\,atoms = n 	imes N_A 	imes 	ext{atoms per molecule}$ where $N_A$ is Avogadro's number ($6.022 	imes 10^{23}$).
  Key Point: For diatomic molecules ($O_2$, $N_2$, $H_2$), multiply the number of moles by $2 	imes N_A$. For Li (monatomic), multiply by $N_A$.
 3. Compare the number of atoms for each substance to find the largest value.
  Result: Let's calculate for each option:

* **Option A (1 g of $O_2$)**: Molar mass of $O_2$ = 32 g/mol.  $n = \frac{1}{32}$ mol. Number of atoms $= \frac{1}{32} 	imes 2 	imes 6.022 	imes 10^{23} \approx 3.76 	imes 10^{22}$ atoms.
* **Option B (1 g of $N_2$)**: Molar mass of $N_2$ = 28 g/mol.  $n = \frac{1}{28}$ mol. Number of atoms $= \frac{1}{28} 	imes 2 	imes 6.022 	imes 10^{23} \approx 4.30 	imes 10^{22}$ atoms.
* **Option C (1 g of $H_2$)**: Molar mass of $H_2$ = 2 g/mol.  $n = \frac{1}{2}$ mol. Number of atoms $= \frac{1}{2} 	imes 2 	imes 6.022 	imes 10^{23} = 6.022 	imes 10^{23}$ atoms.
* **Option D (1 g of $Li$)**: Molar mass of $Li$ = 7 g/mol.  $n = \frac{1}{7}$ mol. Number of atoms $= \frac{1}{7} 	imes 6.022 	imes 10^{23} \approx 8.60 	imes 10^{22}$ atoms.

Option C has the largest number of atoms.
  Verification:  Comparing the calculated number of atoms, 1 g of $H_2$ has significantly more atoms than the others.
 ## OPTION ANALYSIS
A :x:  $3.76 	imes 10^{22}$ atoms
B :x: $4.30 	imes 10^{22}$ atoms
C ✓: $6.022 	imes 10^{23}$ atoms
D :x: $8.60 	imes 10^{22}$ atoms
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Forgetting to multiply by 2 for diatomic molecules: Remember that $O_2$, $N_2$, and $H_2$ are diatomic, meaning they have two atoms per molecule.
  2.Incorrectly calculating molar mass: Double-check the periodic table for the correct atomic masses.
 **Quick Tips**:
  1.Focus on the molar mass: The lower the molar mass, the more moles you'll have for a given mass, leading to a larger number of atoms.
  2.Remember Avogadro's number: It's the key to converting moles to the number of atoms.

Mr. S M HUSSAIN · Answer

1 g of $O_2$

Mr. S M HUSSAIN · Answer

1 g of $N_2$

Mr. S M HUSSAIN · Answer

The correct Option is "1 g of $H_2$"

Mr. S M HUSSAIN · Answer

1 g of $Li$

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