What is the total vapor pressure of NaCl solution at 25C with mole fraction of 0.01? (X_NaCl = 0.01)
Thanks!!
PH2O soln = Xo(solvent)*Po(solvent)
Po solvent = normal vapor pressure of solvent at 25 C.
So since the vapor pressure of water is 23.8 torr
p = 0.99 * 23.8
p = 23.6
is this right?
Looks good to me.
Thanks for your help!!
To find the total vapor pressure of a NaCl solution with a mole fraction of 0.01 at 25°C, we can use Raoult's law.
Raoult's law states that the vapor pressure of a component in a solution is equal to the product of the vapor pressure of the pure component and its mole fraction in the solution.
In this case, we have a NaCl solution with a mole fraction of 0.01, which means that the mole fraction of water (assuming it's a water-based solution) is 0.99.
First, we need to know the vapor pressure of water at 25°C. The vapor pressure of water at this temperature is approximately 23.76 mmHg.
Now, we can use Raoult's law to find the vapor pressure of NaCl in the solution:
P_NaCl = P_water * X_NaCl
P_NaCl = 23.76 mmHg * 0.01
P_NaCl ≈ 0.238 mmHg
Therefore, the total vapor pressure of the NaCl solution at 25°C with a mole fraction of 0.01 is approximately 0.238 mmHg.