Respuesta :
Explanation:
The given data is as follows.
Mass of [tex]O_{2}[/tex] = 1.15 g
Mass of [tex]N_{2}[/tex] = 1.55 g
Therefore, moles of oxygen present will be as follows.
No. of moles of [tex]O_{2}[/tex] = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{1.15 g}{32 g/mol}[/tex]
= 0.035 mol
No. of moles of [tex]N_{2}[/tex] = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{1.55 g}{28.02 g/mol}[/tex]
= 0.055 mol
Hence, total no. of moles = moles of [tex]O_{2}[/tex] + moles of [tex]N_{2}[/tex]
= (0.035 + 0.055) mol
= 0.09 mol
Now, it is known that at STP volume is 22.4 L/mol. Hence, volume of the gas sample at STP for 0.09 moles will be as follows.
[tex]0.09 mol \times 22.4 L/mol[/tex]
= 2.01 L
Thus, we can conclude that volume of the given gas sample is 2.01 L.
