At STP,32 g of O_2 contains 6.022 times 10^{ | vedclass

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

Question

(A) The molar mass of $O_2$ is $32 \ g/mol$. Thus,$32 \ g$ of $O_2$ corresponds to $1 \ mol$ of $O_2$ molecules,which equals $6.022 	imes 10^{23}$ mo

Vedclass · Accepted Answer

$6.022 	imes 10^{23}$

Vedclass · Answer

(A) The molar mass of $O_2$ is $32 \ g/mol$. Thus,$32 \ g$ of $O_2$ corresponds to $1 \ mol$ of $O_2$ molecules,which equals $6.022 	imes 10^{23}$ molecules.Similarly,the atomic mass of sulfur $(S)$ is $32 \ g/mol$. Therefore,$32 \ g$ of sulfur corresponds to $1 \ mol$ of sulfur atoms.Since $1 \ mol$ of any substance contains $6.022 	imes 10^{23}$ particles,$32 \ g$ of sulfur contains $6.022 	imes 10^{23}$ atoms.

At $STP$,$32 \ g$ of $O_2$ contains $6.022 \times 10^{23}$ molecules. How many atoms does $32 \ g$ of sulfur $(S)$ contain under the same conditions?

Similar Questions

How many moles of carbon atoms are present in $3.6 \ kg$ of carbon?

The ratio of the number of oxygen atoms $(O)$ in $16.0 \ g$ ozone $(O_3)$,$28.0 \ g$ carbon monoxide $(CO)$,and $16.0 \ g$ oxygen $(O_2)$ is (Atomic mass: $C = 12$,$O = 16$ and Avogadro's constant $N_A = 6.0 \times 10^{23} \ mol^{-1}$)

Find the number of water molecules in $1 \ mL$ of water vapours at $STP$?

How many electrons are present in $2 \times 10^{-3} \ mol$ of ${}_{8}^{18}O^{2-}$?

What is the number of carbon atoms present in $0.35 \ mol$ of $C_6H_{12}O_6$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module