If we take 44 g of CO_2 and 14 g of N_2,wha | vedclass

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(C) Step $1$: Calculate the number of moles of $CO_2$ $(n_{CO_2} = \frac{44 \ g}{44 \ g/mol} = 1 \ mol)$.Step $2$: Calculate the number of moles of $N

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$2/3$

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(C) Step $1$: Calculate the number of moles of $CO_2$ $(n_{CO_2} = \frac{44 \ g}{44 \ g/mol} = 1 \ mol)$.Step $2$: Calculate the number of moles of $N_2$ $(n_{N_2} = \frac{14 \ g}{28 \ g/mol} = 0.5 \ mol)$.Step $3$: Calculate the mole fraction of $CO_2$ $(X_{CO_2} = \frac{n_{CO_2}}{n_{CO_2} + n_{N_2}} = \frac{1}{1 + 0.5} = \frac{1}{1.5} = \frac{1}{3/2} = 2/3)$.

If we take $44 \ g$ of $CO_2$ and $14 \ g$ of $N_2$,what will be the mole fraction of $CO_2$ in the mixture?

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