10 g each of CO_2,NH_3,and O_2 were taken in | vedclass

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

Question

(D) The number of atoms is calculated as: $	ext{Number of atoms} = 	ext{moles} 	imes N_A 	imes 	ext{atomicity}$.$1$. For $CO_2$: $	ext{moles} =

Vedclass · Accepted Answer

$NH_3 > CO_2 > O_2$

Vedclass · Answer

(D) The number of atoms is calculated as: $	ext{Number of atoms} = 	ext{moles} 	imes N_A 	imes 	ext{atomicity}$.$1$. For $CO_2$: $	ext{moles} = \frac{10}{44} \approx 0.227$. $	ext{Atomicity} = 3$. $	ext{Atoms} = 0.227 	imes 3 	imes N_A = 0.681 N_A$.$2$. For $NH_3$: $	ext{moles} = \frac{10}{17} \approx 0.588$. $	ext{Atomicity} = 4$. $	ext{Atoms} = 0.588 	imes 4 	imes N_A = 2.352 N_A$.$3$. For $O_2$: $	ext{moles} = \frac{10}{32} = 0.3125$. $	ext{Atomicity} = 2$. $	ext{Atoms} = 0.3125 	imes 2 	imes N_A = 0.625 N_A$.Comparing the values: $2.352 N_A (NH_3) > 0.681 N_A (CO_2) > 0.625 N_A (O_2)$.Thus,the decreasing order is $NH_3 > CO_2 > O_2$.

$10 \ g$ each of $CO_2$,$NH_3$,and $O_2$ were taken in three separate flasks. What is the correct decreasing order of the number of atoms?

Similar Questions

If $3.01 \times 10^{20}$ molecules are removed from $98 \ mg$ of $H_2SO_4$,then the number of moles of $H_2SO_4$ left are:

The mass of $10^{21}$ atoms of $Cu$ is ..... grams.

Calculate the total number of atoms in $5.6 \ L$ of $SO_2$ gas at $NTP$.

The number of $H$ atoms in $27 \ g$ of water is .........

The number of moles of electrons in $1 \ kg$ of electrons is .......

Vedclass Test Series

Exam Paper Generator

Online Exam Module