The ratio of masses of oxygen and nitrogen in | vedclass

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

Question

(B) Let the mass of $O_2$ be $w$ and the mass of $N_2$ be $4w$.The molar mass of $O_2$ is $32 \ g/mol$ and the molar mass of $N_2$ is $28 \ g/mol$.The

Vedclass · Accepted Answer

$7 : 32$

Vedclass · Answer

(B) Let the mass of $O_2$ be $w$ and the mass of $N_2$ be $4w$.The molar mass of $O_2$ is $32 \ g/mol$ and the molar mass of $N_2$ is $28 \ g/mol$.The number of molecules is proportional to the number of moles $(n = \frac{	ext{mass}}{	ext{molar mass}})$.Number of molecules of $O_2$ $(N_{O_2})$ = $\frac{w}{32} 	imes N_A$Number of molecules of $N_2$ $(N_{N_2})$ = $\frac{4w}{28} 	imes N_A$Ratio $\frac{N_{O_2}}{N_{N_2}} = \frac{w/32}{4w/28} = \frac{w}{32} 	imes \frac{28}{4w} = \frac{28}{128} = \frac{7}{32}$.

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:

Similar Questions

Which of the following contains the maximum number of atoms?

The number of moles in $620 \ g$ of sodium oxide is ..... moles.

Maximum number of molecules will be in $:-$

Which of the following contains $6.02 \times 10^{22}$ molecules?

Calculate the number of valence electrons in $4.2 \, g$ of $N^{3-}$ ions.

Vedclass Test Series

Exam Paper Generator

Online Exam Module