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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$2/3$

Vedclass · Answer

(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?

Similar Questions

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

$5.6 \, L$ of oxygen at $STP$ contains .............

What is a mole?

Number of moles and number of molecules in $1.4187 \ L$ of $SO_2$ at $STP$ respectively are:

Weight of $112 \, mL$ of oxygen at $NTP$ on liquefaction would be .............. $g$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module