a solution is made by mixing 38.0 ml of ethanol C2H6O, 62.0mL of water. assuming ideal behavior, what is the vapor pressure of the solution at 20 C

Question

a solution is made by mixing 38.0 ml of ethanol C2H6O, 62.0mL of water. assuming ideal behavior, what is the vapor pressure of the solution at 20 C

Answer 1

To calculate the vapor pressure of the solution at 20°C, we need to use Raoult's law. Raoult's law states that the vapor pressure of a component in a solution is proportional to its mole fraction.

Step 1: Find the mole fraction of ethanol (C2H6O):
To find the mole fraction (X) of ethanol, we need to convert the volume of ethanol and water into moles.

Given:
Volume of ethanol (Vethanol) = 38.0 mL
Volume of water (Vwater) = 62.0 mL
Total volume (Vtotal) = Vethanol + Vwater = 38.0 mL + 62.0 mL = 100.0 mL = 100.0 cm^3

Now, we can convert the volumes into liters:
Vethanol = 38.0 cm^3 * (1 mL / 1 cm^3) * (1 L / 1000 mL) = 0.038 L
Vwater = 62.0 cm^3 * (1 mL / 1 cm^3) * (1 L / 1000 mL) = 0.062 L

To calculate the moles of ethanol and water, we'll use their respective densities:
Density of ethanol (ρethanol) = 0.789 g/mL
Density of water (ρwater) = 1.00 g/mL

Mass of ethanol (methanol) = Vethanol * ρethanol = 0.038 L * 0.789 g/mL = 0.029982 g
Mass of water (mwater) = Vwater * ρwater = 0.062 L * 1.00 g/mL = 0.0620 g

Now, we can calculate the moles:
Molar mass of ethanol (Methanol) = 46.07 g/mol
Molar mass of water (Mwater) = 18.02 g/mol

Moles of ethanol (nethanol) = mass of ethanol / molar mass of ethanol = 0.029982 g / 46.07 g/mol ≈ 0.000651 mol
Moles of water (nwater) = mass of water / molar mass of water = 0.0620 g / 18.02 g/mol ≈ 0.003439 mol

Step 2: Calculate the mole fraction (X) of ethanol:
The mole fraction (X) of ethanol is given by the ratio of moles of ethanol to the total moles of the solution.

Xethanol = nethanol / (nethanol + nwater)
= 0.000651 mol / (0.000651 mol + 0.003439 mol)
≈ 0.159

Step 3: Calculate the vapor pressure (P) using Raoult's law:
According to Raoult's law, the vapor pressure (P) of the solution is proportional to the mole fraction of ethanol, multiplied by the vapor pressure of pure ethanol (Pethanol). At 20°C, the vapor pressure of pure ethanol is 44.6 mmHg.

P = Xethanol * Pethanol
= 0.159 * 44.6 mmHg
≈ 7.09 mmHg

Therefore, the vapor pressure of the solution at 20°C, assuming ideal behavior, is approximately 7.09 mmHg.

Answer 2

To calculate the vapor pressure of a solution, we can use Raoult's law, which states that the vapor pressure of a component in an ideal solution is proportional to its mole fraction in the solution.

First, let's find the mole fraction of ethanol (C2H6O) and water in the solution.

Moles of ethanol (C2H6O):
To find the number of moles, we need to know the molar mass of ethanol, which is 46.07 g/mol.

mole of ethanol = (volume of ethanol in mL / 1000) * (density of ethanol / molar mass of ethanol)

density of ethanol is approximately 0.789 g/mL.

mole of ethanol = (38.0 mL / 1000) * (0.789 g/mL / 46.07 g/mol)

Repeat the same steps to find the moles of water (H2O):
mole of water = (62.0 mL / 1000) * (1.00 g/mL of water / 18.02 g/mol)

Next, we can calculate the mole fractions:
moles total = moles of ethanol + moles of water

mole fraction of ethanol = moles of ethanol / moles total
mole fraction of water = moles of water / moles total

Now, we can use Raoult's law to find the vapor pressure of the solution at 20°C.

The vapor pressure of ethanol at 20°C is 44.6 mmHg and the vapor pressure of water at 20°C is 17.5 mmHg.

vapor pressure of the solution = (mole fraction of ethanol) * (vapor pressure of ethanol) + (mole fraction of water) * (vapor pressure of water)

Substitute the values into the formula:
vapor pressure of the solution = (mole fraction of ethanol) * 44.6 mmHg + (mole fraction of water) * 17.5 mmHg

Calculate the values:
vapor pressure of the solution = (mole fraction of ethanol) * 44.6 mmHg + (mole fraction of water) * 17.5 mmHg

Finally, plug in the calculated values of the mole fractions and compute the vapor pressure of the solution.

Answer 3

mols C2H5OH = grams/molar mass. Use density to convert mL to grams.

mols H2O = grams/molar mass. Use density to convert mL to grams.
Then XC2H5OH = nC2H5OH/total mols
XH2O = nH2O/total mols

pH2O = XH2O x P^o_{H2O,/sub>

pC2H5OH = XC2H5OH x P^o_C2H5OH

Ptotal = pC2H5OH + pH2O}