Question

how many nitrogen atoms are present in 32 dm^3 of nitrogen gas at stp​

Answer 1

Answer:

Explanation:

There are 6.022*10^23 molecules present in 22400cm^3 gas.  

Therefore the no. of molecules present in 22.4cm^3=  

6.022*10^23 * 22.4cm^3 divide by:  

224000 cm^3  

Hence we get the ans as 6.022*10^20 molecules.

Answer 2

Answer:

Nitrogen atoms are present in 32 $dm^{3}$ of nitrogen gas at STP​ is 8.599× $10^{23}$atoms

Explanation:

1 cubic decimeter equal to one liter

so we can say that 32 $dm^{3}$ is equal to 32 L.

At STP (standard temperature and pressure), 1 mole of a substance is equivalent to 22.4 L.

that is 22.4 L of nitrogen gas  →  1 mole

1 L of nitrogen gas contains → $\frac{1}{22.4}$ moles of nitrogen

therefore, 32 L of  nitrogen gas contain  →  $\frac{1}{22.4}$ × $32$ =  1.428 moles

Number of atoms = number of moles × Avogadro number

                              = 1.428  ×  6.022×$10^{23}$ = 8.599 × $10^{23}$ atoms