Respuesta :
Answer:- mole fraction of nitrogen is 0.626 and mole fraction of oxygen is 0.374.
partial pressure of nitrogen is 0.687 atm and partial pressure of oxygen is 0.410 atm.
Solution:- Moles of nitrogen = 1.25 g x (1mol/28g) = 0.0446 mol
moles of oxygen = 0.85 g x (1mol/32g) = 0.0266 mol
Total moles of gases in the container = 0.0446 + 0.0266 = 0.0712
mole fraction of a gas = moles of gas/total moles of the gases
so, mole fraction of nitrogen = 0.0446/0.0712 = 0.626
mole fraction of oxygen = 0.0266/0.0712 = 0.374
Volume of the container is 1.55 L and the temperature is 18 degree C that is 18 + 273 = 291 K
From ideal gas law equation, PV = nRT
P = nRT/V
Partial pressure of nitrogen = (0.0446 x 0.0821 x 291)/1.55 = 0.687 atm
and the partial pressure of oxygen = (0.0266 x 0.0821 x 291)/1.55 = 0.410 atm
Moles are calculated as the ratio of given mass to the molar mass. The mole fraction and partial pressure of each component in the mixture of gases will be:
Nitrogen
- Mole fraction = 0.626
- Partial Pressure = 0.687 atm
Oxygen
- Mole fraction = 0.374
- Partial Pressure = 0.410 atm
Moles are the base units of a substance. It is calculated as the ratio of given mass to the molar mass.
Now, the moles fraction of nitrogen will be:
- [tex]\begin{aligned}\text{Mole}&=1.25\times\dfrac{1 \text{mol}}{28 \text g}\\\\&=0.0446 \text{mol}\end[/tex]
Similarly, moles of oxygen will be:
- [tex]\begin{aligned}\text{Mole}&=0.85\times\dfrac{1 \text{mol}}{32 \text g}\\\\&=0.0266 \text{mol}\end[/tex]
Now, Total moles of gases = 0.0446 + 0.0266 = 0.0712
Also,
- [tex]\text{Mole fraction of gas}&=\dfrac{\text{moles of gas}}{\text{total moles of gases}}[/tex]
Mole fraction of nitrogen:
- [tex]\begin{aligned}\text{Mole fraction of gas}&=\dfrac{\text{moles of Nitrogen}}{\text{total moles of gases}}\\\\\text{Mole fraction of Nitrogen}&=\dfrac{0.0446}{0.0712}\\\\&=0.626\end{aligned}[/tex]
Similarly, for oxygen:
- [tex]\begin{aligned}\text{Mole fraction of gas}&=\dfrac{\text{moles of Oxygen}}{\text{total moles of gases}}\\\\\text{Mole fraction of Oxygen}&=\dfrac{0.0266}{0.0712}\\\\&=0.374\end{aligned}[/tex]
Also, given that,
- Volume of container = 1.55 L
- Temperature = 18-degree celcius
- Ideal gas equation = PV = nRT
Substituting the values:
Partial pressure of nitrogen:
- [tex]\begin{aligned}\text{Partial pressure of nitrogen\;(P)}&=\frac{\text{nRT}}{\text{V}}\\&=\frac{(0.0446\times0.0821\times291)}{1.55}\\&=0.687\;\text{atm}\end{aligned}[/tex]
Partial Pressure of Oxygen:
- [tex]\begin{aligned}\text{Partial pressure of oxygen\;(P)}&=\frac{\text{nRT}}{\text{V}}\\&=\frac{(0.0266\times0.0821\times291)}{1.55}\\&=0.410\;\text{atm}\end{aligned}[/tex]
Therefore, from the the ideal gas law equation and mole fraction, we can determine the number of moles and partial pressure of gases.
To know more about partial pressure and moles, refer to the following link:
https://brainly.com/question/13785323?referrer=searchResults
