If a mixture of 28 , g of Nitrogen (N_2),4 , | vedclass

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(C) The ideal gas law for a mixture is given by $PV = n_{total}RT$,where $n_{total} = \sum \frac{m_i}{M_i}$.First,calculate the number of moles for ea

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$2 \, g/litre$

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(C) The ideal gas law for a mixture is given by $PV = n_{total}RT$,where $n_{total} = \sum \frac{m_i}{M_i}$.First,calculate the number of moles for each gas:$n_{N_2} = \frac{28 \, g}{28 \, g/mol} = 1 \, mol$$n_{H_2} = \frac{4 \, g}{2 \, g/mol} = 2 \, mol$$n_{He} = \frac{8 \, g}{4 \, g/mol} = 2 \, mol$Total moles $n_{total} = 1 + 2 + 2 = 5 \, mol$.Total mass $m_{total} = 28 + 4 + 8 = 40 \, g = 0.04 \, kg$.Using the ideal gas equation $PV = n_{total}RT$,we find the volume $V = \frac{n_{total}RT}{P}$.$V = \frac{5 	imes 8.3 	imes 400}{8.3 	imes 10^5} = \frac{2000}{10^5} = 0.02 \, m^3$.Density $ho = \frac{m_{total}}{V} = \frac{0.04 \, kg}{0.02 \, m^3} = 2 \, kg/m^3$.Since $1 \, kg/m^3 = 1 \, g/litre$,the density is $2 \, g/litre$.

If a mixture of $28 \, g$ of Nitrogen $(N_2)$,$4 \, g$ of Hydrogen $(H_2)$,and $8 \, g$ of Helium $(He)$ is contained in a vessel at temperature $400 \, K$ and pressure $8.3 \times 10^5 \, Pa$,the density of the mixture will be:

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