A solution of sodium chloride in water has a vapor pressure of 18.5 torr at 25°C.
What is the mole fraction of NaCl in this solution? What would be the vapor pressure of this solution at 48°C? The vapor pressure of pure water is 23.8 torr at 25°C and 83.7 torr at 48°C.
I've tried this as well as went to my tutor, but we couldn't get it. Can someon ehelp me? I need it by tonight
And if there seems to be confusion why did you not post the methods you had tried and the work. That way we could look them over. Sometimes third persons can find the error.
Psoln = XsolventPo
18.5 = X*23.8 should give you the mole fraction OF WATER. The mole fraction of NaCl will be 1 minus that (and my guess is that's where you and your tutor missed the boat). For the second part,
Psoln = Xwater*83.7
njk
Sure, I'd be happy to help you solve this problem.
To find the mole fraction of NaCl in the solution, we need to calculate the number of moles of NaCl and water separately.
First, let's find the mole fraction of NaCl at 25°C.
1. We know the vapor pressure of the solution, P_solution, is 18.5 torr.
2. The vapor pressure of pure water, P_water, is 23.8 torr.
3. The vapor pressure of pure NaCl, P_NaCl, is negligible compared to P_water, so we can ignore it in our calculations.
4. According to Raoult's law, the vapor pressure of a solution is equal to the product of the mole fraction of the solvent (water) and the vapor pressure of pure water at that temperature.
Therefore, we can use the following equation: P_solution = x_water * P_water, where x_water is the mole fraction of water in the solution.
To find x_water, rearrange the equation: x_water = P_solution / P_water.
Substituting the given values, we have:
x_water = 18.5 torr / 23.8 torr ≈ 0.775, or 77.5%
This means that water makes up approximately 77.5% of the solution at 25°C, and since NaCl is the solute, its mole fraction (x_NaCl) can be calculated as 1 - x_water = 1 - 0.775 ≈ 0.225, or 22.5%.
Now, let's find the vapor pressure of the solution at 48°C.
1. We know the vapor pressure of pure water, P_water, at 48°C is 83.7 torr.
2. We already know the mole fraction of water, x_water = 0.775, from the previous calculation at 25°C.
3. We need to find the mole fraction of NaCl at 48°C, denoted as x_NaCl_48.
According to Raoult's law, the vapor pressure of the solution at 48°C is given by P_solution_48 = x_water_48 * P_water_48, where x_water_48 is the mole fraction of water at 48°C.
To find x_water_48, rearrange the equation: x_water_48 = P_solution_48 / P_water_48.
Substituting the given values, we get:
x_water_48 = (18.5 torr / 23.8 torr) * (83.7 torr / 23.8 torr) ≈ 0.615, or 61.5%
This means that water makes up approximately 61.5% of the solution at 48°C. Since NaCl is the solute, its mole fraction (x_NaCl_48) can be calculated as 1 - x_water_48 = 1 - 0.615 ≈ 0.385, or 38.5%.
Therefore, the mole fraction of NaCl in the solution at 48°C is approximately 0.385, or 38.5%.
I hope this explanation helps! Let me know if you have any further questions.