In which of the following pairs do 1 text{ g} | vedclass

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(C) The number of molecules in a given mass is calculated as: $	ext{Number of molecules} = \frac{	ext{Mass}}{	ext{Molar Mass}} 	imes N_A$,where $N

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$(III)$ and $(IV)$

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(C) The number of molecules in a given mass is calculated as: $	ext{Number of molecules} = \frac{	ext{Mass}}{	ext{Molar Mass}} 	imes N_A$,where $N_A$ is Avogadro's number $(6.022 	imes 10^{23} 	ext{ mol}^{-1})$.
For $1 	ext{ g}$ of each substance to have the same number of molecules,their molar masses must be equal.
Let's calculate the molar masses:
$(I) \, N_2O = (2 	imes 14) + 16 = 44 	ext{ g/mol}$; $CO = 12 + 16 = 28 	ext{ g/mol}$. (Not equal)
$(II) \, N_2 = 28 	ext{ g/mol}$; $CO_2 = 12 + (2 	imes 16) = 44 	ext{ g/mol}$. (Not equal)
$(III) \, N_2 = 28 	ext{ g/mol}$; $CO = 28 	ext{ g/mol}$. (Equal)
$(IV) \, N_2O = 44 	ext{ g/mol}$; $CO_2 = 44 	ext{ g/mol}$. (Equal)
Thus,pairs $(III)$ and $(IV)$ contain the same number of molecules for $1 	ext{ g}$ of each substance.

In which of the following pairs do $1 \text{ g}$ of each contain the same number of molecules?
$(I)\, N_2O$ and $CO$ $(II)\,N_2$ and $CO_2$
$(III)\, N_2$ and $CO$ $(IV) \,N_2O$ and $CO_2$

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In which of the following pairs do $1 \text{ g}$ of each contain the same number of molecules? $(I)\, N_2O$ and $CO$ $(II)\,N_2$ and $CO_2$ $(III)\, N_2$ and $CO$ $(IV) \,N_2O$ and $CO_2$