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(B) Step $1$: Calculate the moles of each gas.Moles of $N_2 = \frac{3.01 	imes 10^{24}}{6.02 	imes 10^{23}} = 5 \ mol$.Moles of $O_2 = \frac{32 \ g}

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(B) Step $1$: Calculate the moles of each gas.Moles of $N_2 = \frac{3.01 	imes 10^{24}}{6.02 	imes 10^{23}} = 5 \ mol$.Moles of $O_2 = \frac{32 \ g}{32 \ g/mol} = 1 \ mol$.Total moles $(n)$ = $5 + 1 = 6 \ mol$.Step $2$: Calculate the total volume of the mixture using the ideal gas equation $PV = nRT$.$V = \frac{nRT}{P} = \frac{6 	imes (1/12) 	imes 860}{3}$.$V = \frac{0.5 	imes 860}{3} = \frac{430}{3} \approx 143.33 \ L$.Step $3$: Calculate the total mass of the mixture.Mass of $N_2 = 5 \ mol 	imes 28 \ g/mol = 140 \ g$.Mass of $O_2 = 32 \ g$.Total mass = $140 + 32 = 172 \ g$.Step $4$: Calculate the density $(d = \frac{mass}{volume})$.$d = \frac{172}{143.33} \approx 1.2 \ g/L$.

Calculate the density of a gaseous mixture which consists of $3.01 \times 10^{24}$ molecules of $N_2$ and $32 \ g$ of $O_2$ gas at $3 \ atm$ pressure and $860 \ K$ temperature. (Given: $R = 1/12 \ atm \ L \ mol^{-1} \ K^{-1}$) ............. $g/L$

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