Pvap (mm-Hg) of 100g 24%NaOH in H2O @100C? When Pvap pure H2O @100C is 760mm-Hg
Pls where to start solving this problem? What is the equation? What are the staeps to get to the right answer?
P = XH2O*Po
So you need to calculate mols NaOH and mols H2O, then XH2O and substitute into the above with Po = 760.
Thx! for Mole of H2O: I took 76g since it is 100g of 24%NaOH solution ????
Then,
Mole of H2O: 76/18=4.222
Mole of NaOH: 24/39.997=0.6
Total moles: 4.222+0.6=4.822
X1=4.222/4.822=0.876
P1=0.876x760=665.76 mm Hg
To solve this problem, you need to use Raoult's law, which relates the vapor pressure of a solution to the vapor pressure of its components. The equation is as follows:
Pvap(solution) = Xsolvent * Pvap(solvent)
Here, Pvap(solution) is the vapor pressure of the solution, Xsolvent is the mole fraction of the solvent, and Pvap(solvent) is the vapor pressure of the pure solvent.
Now, let's break down the steps to get to the right answer:
Step 1: Calculate the mole fraction of the solvent:
In this case, the solvent is water (H2O). Since the NaOH is present in a 24% weight by weight solution, it means that 76% of the solution is water. To calculate the mole fraction of the solvent, divide the weight percent of water by its molar mass:
Molar mass of water (H2O) = 18.02 g/mol
Mole fraction of H2O = (0.76 * 100g) / (18.02 g/mol)
Step 2: Determine the vapor pressure of pure water:
Given that the vapor pressure of pure water at 100°C is 760 mmHg, you can use this value as Pvap(solvent). If the temperature were different, you would need to consult a vapor pressure table specific to water.
Step 3: Calculate the vapor pressure of the solution:
Now you can use Raoult's law to determine the vapor pressure of the solution. Multiply the mole fraction of water (Xsolvent) by the vapor pressure of pure water (Pvap(solvent)):
Pvap(solution) = Xsolvent * Pvap(solvent)
Plug in the values obtained in steps 1 and 2 to calculate the answer.
Remember that when using Raoult's law, it is assumed that the solute (NaOH) does not contribute significantly to the vapor pressure and that the solution behaves ideally.