Which of the following contains the maximum n | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) To find the number of atoms,we calculate the number of moles and multiply by the atomicity of the molecule.$1$. For $16 \, g \, O_2$: Moles = $16

Vedclass · Accepted Answer

$2 \, g \, H_2$

Vedclass · Answer

(C) To find the number of atoms,we calculate the number of moles and multiply by the atomicity of the molecule.$1$. For $16 \, g \, O_2$: Moles = $16 / 32 = 0.5 \, mol$. Atoms = $0.5 	imes 2 	imes N_A = 1.0 \, N_A$.$2$. For $14 \, g \, N_2$: Moles = $14 / 28 = 0.5 \, mol$. Atoms = $0.5 	imes 2 	imes N_A = 1.0 \, N_A$.$3$. For $2 \, g \, H_2$: Moles = $2 / 2 = 1.0 \, mol$. Atoms = $1.0 	imes 2 	imes N_A = 2.0 \, N_A$.$4$. For $6 \, g \, I_2$: Moles = $6 / 254 \approx 0.023 \, mol$. Atoms = $0.023 	imes 2 	imes N_A \approx 0.046 \, N_A$.Comparing the values,$2 \, g \, H_2$ contains the maximum number of atoms.

Which of the following contains the maximum number of atoms?

Similar Questions

What is the number of molecules present in $44 \ g$ of $CO_2$?

Which of the following contains the maximum number of atoms?

Which of the following contains the maximum number of molecules?

If Avogadro number $N_A$ is changed from $6.022 \times 10^{23} \ mol^{-1}$ to $6.022 \times 10^{20} \ mol^{-1},$ this would change

Which of the following is the Loschmidt number?

Vedclass Test Series

Exam Paper Generator

Online Exam Module