A gaseous mixture consists of 75.0 mole percent N_2 and 25.0 mole percent O_2 (t

Question

A gaseous mixture consists of 75.0 mole percent N_2 and 25.0 mole percent O_2 (the approximate composition of air). Suppose water is saturated with the gas mixture at 25 degree C and 1.00 atm total pressure, and then the gas is expelled from the water by heating. What is the composition in mole fraction of the gas mixture that is expelled? The solubilities of N_2 and O_2 at 25 degree C and 1.00 atm are 0.0175 g/L H_2O and 0.0393 g/L H_2O, respectively. Mole fraction of N_2 = Mole fraction of O_2 =

Explanation / Answer

The molar mass of N2= 28 g/ mole

The molar mass of O2= 32 g/ mole

Assume that the total volume = 1000 L

Given that;

the solubility of N2= 0.0175 g/ L

the solubility of O2= 0.0393 g/ L

Then amount of N2 and O2 in this 1000 L volume

N2 = 0.0175 g/ L *1000 L

= 17.5 g

O2 = 0.0393 g/ L *1000 L

= 39.3 g

Now number of moles of N2 and O2:

Number of mole s= amount in g / molar mass

Now number of moles of N2 =17.5 g/28.0g/ mole

= 0.625 mole

Now number of moles of O2 =39.3 g/32.0g/ mole

= 1.23 mole

Mole fraction:

Mole fraction is a ratio of mole of solute or solvent to the total number of moles (mole of solute + moles of solvent) or number of mole of solution.

Mole fraction of N2 = 0.625 mole /0.625 mole + 1.23 mole

=0.625 mole / 1.855 mole

= 0.34

Mole fraction of O2 = 1.23 mole /0.625 mole + 1.23 mole

=1.23 mole / 1.855 mole

= 0.66

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.