Answer:
[tex]0.15008\ \text{g}[/tex]
[tex]3.23\times 10^{21}[/tex]
Explanation:
1 mol of nitrogen at STP = 22.4 L = 22400 cc
n = Mol of [tex]N_2[/tex] = [tex]\dfrac{120}{22400}=0.00536\ \text{mol}[/tex]
M = Molar mass of [tex]N_2[/tex] = [tex]28\ \text{g/mol}[/tex]
[tex]N_A[/tex] = Avogadro's number = [tex]6.022\times 10^{23}\ \text{mol}^{-1}[/tex]
Mass of [tex]N_2[/tex] is
[tex]m=nM\\\Rightarrow m=0.00536\times 28\\\Rightarrow m=0.15008\ \text{g}[/tex]
Mass of the nitrogen is [tex]0.15008\ \text{g}[/tex]
Number of molecules is given by
[tex]nN_A=0.00536\times 6.022\times 10^{23}=3.23\times 10^{21}\ \text{molecules}[/tex]
The number of molecules present in it are [tex]3.23\times 10^{21}[/tex]
