What is the vapor pressure of a aqueous solution containing 10 % (by weight) ethylene glycol (62 g/mol) at 25 °C. PH2O = 24.3 torr at 25°C?
23.5 torr
10% w/w means 10 g ethylene glycol per 100 g solution.
mols ethylene glycol = grams/molar mass
10g eth/(90g H2O + 10g eth)
mols in 90g H2O = grams/molar mass
total mols = mols ethylene glycol + mols H2O
Then XH2O = mols H2O/total mols.
Psoln = XH2O*PoH2O
Well, you know what they say, the solution to this problem is quite a d-oh-lution! So, let's dive in, shall we?
To find the vapor pressure of the aqueous solution, we need to calculate the mole fraction of ethylene glycol (C2H6O2) in the solution. Mole fraction, what a mole-nificent term!
To start, we need to convert the given weight percentage to mole fraction. We have 10% ethylene glycol in the solution, which means that we have 10 grams of ethylene glycol per 100 grams of solution.
Now, let's calculate the moles of ethylene glycol:
Moles of ethylene glycol = (10 g / 62 g/mol) = 0.1613 moles
Next, let's calculate the moles of water:
Moles of water = (100 g - 10 g) / 18 g/mol = 5.5556 moles
Now, let's calculate the mole fraction of ethylene glycol:
Mole fraction of ethylene glycol = (0.1613 moles) / (0.1613 moles + 5.5556 moles) = 0.0288
Ah, we're getting some clarity now!
Since the vapor pressure of the solution is directly proportional to the mole fraction of the solute, we can now calculate the vapor pressure of the solution.
Vapor pressure of solution = (0.0288) * (24.3 torr) = 0.70074 torr
Ta-da! The vapor pressure of the aqueous solution containing 10% ethylene glycol at 25 °C is approximately 0.70074 torr.
Hope that helps! If you have any more chemistry questions, feel free to ask.
To find the vapor pressure of the aqueous solution containing 10% ethylene glycol at 25 °C, we can use Raoult's Law, which states that the vapor pressure of a solution is equal to the mole fraction of the solvent times the vapor pressure of the pure solvent.
1. First, let's calculate the mole fraction of the solvent (water). Since the solution contains 10% ethylene glycol, the remaining 90% must be water.
Molar mass of ethylene glycol (C2H6O2) = 2*12.01 + 6*1.008 + 2*16.00 = 62.07 g/mol
Molar mass of water (H2O) = 2*1.008 + 16.00 = 18.02 g/mol
Therefore, the weight of water in the solution = 90% * total weight of solution
Weight of water = 90% * (62 g/mol / 62.07 g/mol) * (18.02 g/mol)
Weight of water = 16.11 g
The weight of the total solution is given as 62 g/mol. Therefore, the weight of ethylene glycol is 10% of 62 g/mol, which is 6.2 g.
2. Next, convert the weights of the solvent and solute into moles using their respective molar masses.
Moles of water = weight of water / molar mass of water
Moles of water = 16.11 g / 18.02 g/mol
Moles of water ≈ 0.894 mol
Moles of ethylene glycol = weight of ethylene glycol / molar mass of ethylene glycol
Moles of ethylene glycol = 6.2 g / 62.07 g/mol
Moles of ethylene glycol ≈ 0.1 mol
3. Calculate the mole fraction of the solvent (water).
Mole fraction of water = moles of water / total moles
Mole fraction of water = 0.894 mol / (0.894 mol + 0.1 mol)
Mole fraction of water ≈ 0.899
4. Finally, use Raoult's Law to find the vapor pressure of the solution.
Vapor pressure of the solution = mole fraction of water * vapor pressure of pure water
Vapor pressure of the solution = 0.899 * 24.3 torr
Vapor pressure of the solution ≈ 21.86 torr (rounded to two decimal places)
Therefore, the vapor pressure of the aqueous solution containing 10% ethylene glycol at 25 °C is approximately 21.86 torr.
To determine the vapor pressure of an aqueous solution containing ethylene glycol, we need to apply Raoult's law. Raoult's law states that the partial pressure of a component in an ideal solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.
First, let's calculate the mole fraction of ethylene glycol in the solution. We can assume that the density of the solution is 1 g/mL.
1. Calculate the mass of the solution:
We know that the solution contains 10% ethylene glycol, which means that 10 grams of ethylene glycol are present in 100 grams of the solution.
2. Calculate the moles of ethylene glycol:
We divide the mass of ethylene glycol by its molar mass (62 g/mol) to find the moles of ethylene glycol in the solution:
Moles of ethylene glycol = 10 g / 62 g/mol
3. Calculate the moles of water:
Since the remaining 90 grams of the solution is water, we can find the moles of water by dividing its mass by its molar mass (18 g/mol):
Moles of water = 90 g / 18 g/mol
4. Calculate the mole fraction of ethylene glycol:
The mole fraction of a component in a solution is equal to the moles of the component divided by the total moles of all components:
Mole fraction of ethylene glycol = Moles of ethylene glycol / (Moles of ethylene glycol + Moles of water)
Now that we have the mole fraction of ethylene glycol, we can calculate the vapor pressure of the solution.
5. Calculate the vapor pressure of ethylene glycol:
Using Raoult's law, we multiply the mole fraction of ethylene glycol by the vapor pressure of pure ethylene glycol (which is assumed to be 24.3 torr) at 25 °C to get the vapor pressure of the solution:
Vapor pressure of ethylene glycol = Mole fraction of ethylene glycol * Vapor pressure of pure ethylene glycol
Therefore, the vapor pressure of the aqueous solution containing 10% ethylene glycol can be obtained by following these steps.