Answer:
Initially [tex]1.51\times 10^{-2} moles[/tex] of nitrogen dioxide were in the container .
Explanation:
Volume of the container at low pressure and at room temperature =[tex]V_1=3.4 L[/tex]
Number of moles in the container = [tex]n_1[/tex]
After more addition of nitrogen gas at the same pressure and temperature.
Volume of the container after addition = [tex]V_2=5.11 L[/tex]
Number of moles in the container after addition=[tex]n_2=2.28\times 10^{-2} mol[/tex]
Applying Avogadro's law:
[tex]\frac{Volume}{Moles}=constant[/tex] (at constant pressure and temperature)
[tex]\frac{V_1}{n_1}=\frac{V_2}{n_2}[/tex]
[tex]n_1=\frac{V_1\times n_2}{V_2}=\frac{3.4 L\times 2.28\times 10^{-2} mol}{5.11 L}[/tex]
[tex]n_1=1.51\times 10^{-2} mol[/tex]
Initially [tex]1.51\times 10^{-2} moles[/tex] of nitrogen dioxide were in the container .