Answer:
[tex]\boxed {\boxed {\sf 17.3 \ L \ F_2}}[/tex]
Explanation:
First, we must convert molecules to moles.
We use Avogadro's Number: 6.022*10²³. This number tells us the amount of particles (atoms, molecules, etc.) in 1 mole of a substance. In this case, it is molecules of F₂
[tex]\frac{ 6.022*10^{23} \ molecules \ F_2}{1 \ mol \ F_2}[/tex]
Multiply by the given number of molecules.
[tex]4.65 *10^{23} \ molecules \ F_2*\frac{ 6.022*10^{23} \ molecules \ F_2}{1 \ mol \ F_2}[/tex]
Flip the fraction so the molecules of fluorine cancel.
[tex]4.65 *10^{23} \ molecules \ F_2*\frac{1 \ mol \ F_2 }{6.022*10^{23} \ molecules \ F_2}[/tex]
[tex]4.65 *10^{23} *\frac{1 \ mol \ F_2 }{6.022*10^{23} }[/tex]
[tex]\frac{4.65 *10^{23} \ mol \ F_2 }{6.022*10^{23} }=0.7721687147 \ mol \ F_2[/tex]
Next, convert the moles to liters. Assuming this is at STP (standard temperature and pressure), there are 22.4 liters in 1 mole of any gas.
[tex]\frac {22.4 \ L \ F_2} {1 \ mol \ F_2}[/tex]
Multiply by the number of moles we calculated.
[tex]0.7721687147 \ mol \ F_2*\frac {22.4 \ L \ F_2} {1 \ mol \ F_2}[/tex]
The moles of fluorine cancel.
[tex]0.7721687147 *\frac {22.4 \ L \ F_2} {1 }[/tex]
[tex]0.7721687147 *\ {22.4 \ L \ F_2} =17.29657921 \ L \ F_2[/tex]
The original measurement has 3 significant figures (4, 6, and 5), so our answer must have the same. For the number we calculated, that is the tenth place. The 9 in the hundredth place tells us to round the 2 up to a 3.
[tex]17.3 \ L \ F_2[/tex]
There are approximately 17.3 liters of fluorine.
