Question

How many liters are occupied by 1.806 x 1024 atoms of helium gas at STP?

Answer 1

Answer : The volume of helium gas occupied are, 67.2 L

Explanation :

First we have to calculate the number of moles of helium gas.

As we know that, 1 mole of gas contains $6.022\times 10^{23}$ number of atoms.

As, $6.022\times 10^{23}$ number of helium atoms present in 1 mole of helium gas

So, $1.806\times 10^{24}$ number of helium atoms present in $\frac{1.806\times 10^{24}}{6.022\times 10^{23}}\times 1=2.999$ mole of helium gas

Now we have to calculate the volume of helium gas occupied.

As we know that at STP, 1 mole of gas occupies 22.4 L volume of gas.

As, 1 mole of helium gas occupies 22.4 L volume of helium gas

So, 2.999 mole of helium gas occupies $2.999\times 22.4L=67.2L$ volume of helium gas

Thus, the volume of helium gas occupied are, 67.2 L