Given the ratio of masses of nitrogen and oxy | vedclass

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(B) The number of molecules $(N)$ is given by the formula: $N = n 	imes N_A$,where $n$ is the number of moles and $N_A$ is the Avogadro constant. Sin

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$32: 7$

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(B) The number of molecules $(N)$ is given by the formula: $N = n 	imes N_A$,where $n$ is the number of moles and $N_A$ is the Avogadro constant. Since $n = \frac{w}{M}$ (where $w$ is the mass and $M$ is the molar mass),we have $N = \frac{w}{M} 	imes N_A$. Therefore,the ratio of the number of molecules of $N_2$ to $O_2$ is: $\frac{N_{N_2}}{N_{O_2}} = \frac{w_{N_2}}{M_{N_2}} 	imes \frac{M_{O_2}}{w_{O_2}} = \left( \frac{w_{N_2}}{w_{O_2}} ight) 	imes \left( \frac{M_{O_2}}{M_{N_2}} ight)$. Given $\frac{w_{N_2}}{w_{O_2}} = \frac{4}{1}$,$M_{N_2} = 28 \ g/mol$,and $M_{O_2} = 32 \ g/mol$. Substituting these values: $\frac{N_{N_2}}{N_{O_2}} = \frac{4}{1} 	imes \frac{32}{28} = \frac{4 	imes 8}{7} = \frac{32}{7}$. Thus,the ratio is $32: 7$.

Given the ratio of masses of nitrogen and oxygen in a particular gaseous mixture is $4: 1$. Calculate the ratio of the number of their molecules.

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