Which of the following contains 6.02 times 10 | vedclass

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Question

(A) The number of molecules is calculated using the formula: $	ext{Number of molecules} = \frac{	ext{Given mass}}{	ext{Molar mass}} 	imes N_A$. Gi

Vedclass · Accepted Answer

$2.8 \ g \ 	ext{of} \ N_2$

Vedclass · Answer

(A) The number of molecules is calculated using the formula: $	ext{Number of molecules} = \frac{	ext{Given mass}}{	ext{Molar mass}} 	imes N_A$. Given $N_A = 6.02 	imes 10^{23} \ 	ext{molecules/mol}$. We need $6.02 	imes 10^{22}$ molecules,which is $0.1 \ 	ext{mol}$. For $A$: $2.8 \ g \ N_2$ $(M = 28 \ g/mol)$ = $0.1 \ 	ext{mol}$. Since $N_2$ is a molecule,it contains $0.1 	imes 6.02 	imes 10^{23} = 6.02 	imes 10^{22}$ molecules. For $B$: $1.6 \ g \ O_2$ $(M = 32 \ g/mol)$ = $0.05 \ 	ext{mol}$. For $C$: $Na$ is an atom,not a molecule. For $D$: $1.4 \ g \ O_2$ $(M = 32 \ g/mol)$ = $0.04375 \ 	ext{mol}$. Thus,$A$ is the correct answer.

Which of the following contains $6.02 \times 10^{22}$ molecules?

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