19. (Ch5) In a mixture of N_(2)(g) and O_(2)(g) the partial pressure of N_(2)(g) is 0.815 atm . If the total pressure of the mixture is 1.085 atm, the mole fraction of N_(2) in the mixture is a. 0.249 b. 0.27 o c. 0.712 d. 0.815 e. 0.751

To find the mole fraction of \(N_2\) in the mixture, we can use the partial pressure of \(N_2\) and the total pressure of the mixture.<br /><br />The mole fraction (\(X_{N_2}\)) is given by the ratio of the partial pressure of \(N_2\) to the total pressure:<br /><br />\[X_{N_2} = \frac{P_{N_2}}{P_{total}}\]<br /><br />Given:<br />- Partial pressure of \(N_2\) (\(P_{N_2}\)) = 0.815 atm<br />- Total pressure of the mixture (\(P_{total}\)) = 1.085 atm<br /><br />Substituting the values:<br /><br />\[X_{N_2} = \frac{0.815 \text{ atm}}{1.085 \text{ atm}}\]<br /><br />\[X_{N_2} \approx 0.751\]<br /><br />Therefore, the mole fraction of \(N_2\) in the mixture is approximately 0.751.<br /><br />So, the correct answer is:<br /><br />e. 0.751