Question

N\(_2\) exerts partial pressure of 7.698 bar in 1L water at 298K. What is the mole fraction of N\(_2\) in the solution? Given that N\(_2\) = 76.48K bar.

• A. 0.1
• B. 0.2
• C. 0.3
• D. 0.4

Hint:

To find the mole fraction using Raoult's Law, divide the partial pressure of the component by its vapor pressure at the given temperature.

Answer 1

The Correct Option is A

We can use Raoult's Law to calculate the mole fraction of N\(_2\) in the solution. According to Raoult's Law, the mole fraction of a component in a solution is given by: \[ X = \frac{P}{P^0} \] Where: - \(X\) is the mole fraction, - \(P\) is the partial pressure of the component in the solution, - \(P^0\) is the vapor pressure of the pure component. From the question: - The partial pressure of N\(_2\) is given as 7.698 bar. - The vapor pressure of pure N\(_2\) at 298K is 76.48K bar. Substituting the given values into the equation: \[ X_{\text{N}_2} = \frac{7.698}{76.48} = 0.1 \] Thus, the mole fraction of N\(_2\) in the solution is 0.1.