Question

How many atoms of hydrogen are in 67.2 L of H2 at STP?

Answer 1

Answer:

One mole of any substance contains Avogadro's number of atoms = 6.022 x 10^23 atoms. So multiply number of moles x number of atoms/mole = 1.8066 x 10^24 atoms of H2.

Answer 2

Answer:

The number of hydrogen atoms presents in $67.2L$ of $H_{2}$ gas at STP measured is $3.6\times 10^{24}$ atoms.

Explanation:

Data given,

The volume of $H_{2}$ gas = $67.2L$

The number of hydrogen atoms in $67.2L$ of $H_{2}$ gas at STP  =?

As we know,

• The volume of gas at STP ( standard temperature and pressure ) = $22.4L/mol$

Now the number of moles = $\frac{V_{H_{2} } }{V_{STP} }$ = $\frac{67.2}{22.4}$ = $3mol$

Now, we know that,

• $1mole$ = $6.022\times 10^{23}$ atoms
• $3 moles$ = $3\times 6.022\times 10^{23}$ atoms = $18.066\times 10^{23}$ atoms

As we know, in $H_{2}$ gas, there are 2 atoms of hydrogen.

Thus,

• The number of hydrogen atoms presents in $67.2L$ of $H_{2}$ gas at STP = $2\times 18.066\times 10^{23}$ atoms = $3.6\times 10^{24}$ atoms.