Question

Water has a vapor pressure of 18.0 torr and a density of 0.997 g/mL. Methanol (CH3 OH) has a vapor pressure of 126.8 torr and a density of 0.791 g/mL. Calculate the vapor pressure (in torr) of a solution containing 525 mL of water and 255 mL of methanol. Assume this solution behaves ideally

Answer 1

The vapor pressure of a solution containing 525 mL of water and 255 mL of methanol can be calculated by applying Raoult's Law. First, calculate the moles of each component using their densities and molar masses, then calculate the mole fractions. Using these, the solution's vapor pressure can be found by multiplying each substance's vapor pressure with its mole fraction and summing the results.

Step-by-step explanation:

The question asks us to calculate the vapor pressure of a solution containing 525 mL of water and 255 mL of methanol. To solve this, we need to use Raoult's Law, which is applicable for ideal solutions. According to Raoult's Law, the vapor pressure of a solution is the sum of the vapor pressures of each component multiplied by its mole fraction in the solution (P = P1χ1 + P2χ2).

First, we calculate the moles of water and methanol using their given densities and molecular weights (water = 18.01528 g/mol, methanol = 32.04 g/mol).

Moles of water = (density of water * volume of water) / molecular weight of water

Moles of methanol = (density of methanol * volume of methanol) / molecular weight of methanol

Next, we find the total moles in the solution and then calculate the mole fractions of water (χwater) and methanol (χmethanol).

Mole fraction of water = moles of water / total moles

Mole fraction of methanol = moles of methanol / total moles

Finally, we apply Raoult's Law to find the vapor pressure of the solution:

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

Using the given vapor pressures and the above calculations, we can find the vapor pressure of the mixed solution.

Answer 2

The vapor pressure of an ideal solution containing 525 mL of water and 255 mL of methanol is 37.4 torr. This is calculated by determining the mole fractions of water and methanol, and applying Raoult's Law.

Step-by-step explanation:

To calculate the vapor pressure of a solution containing 525 mL of water and 255 mL of methanol, we use Raoult's Law which states that the vapor pressure of an ideal solution is directly proportional to the mole fraction of the solvent and the vapor pressure of the pure solvents. To find the mole fractions, we need to know the number of moles of each component in the solution.

Firstly, let's calculate the number of moles of water (H2O) and methanol (CH3OH). The molecular weight of water is approximately 18.015 g/mol, and that of methanol is about 32.04 g/mol.

The number of moles of water: (525 mL) x (0.997 g/mL) / (18.015 g/mol) = 29.1 moles of water.
The number of moles of methanol: (255 mL) x (0.791 g/mL) / (32.04 g/mol) = 6.3 moles of methanol.

Now, we can calculate the mole fractions (X):

• Xwater = moles of water / total moles = 29.1 / (29.1 + 6.3) = 0.822
• Xmethanol = moles of methanol / total moles = 6.3 / (29.1 + 6.3) = 0.178

Finally, we apply Raoult's Law to find the vapor pressure of the solution:

• Psoln = Xwater x Pwater + Xmethanol x Pmethanol
• Psoln = (0.822 x 18.0 torr) + (0.178 x 126.8 torr)
• Psoln = 14.8 torr + 22.6 torr
• Psoln = 37.4 torr

The vapor pressure of the solution, assuming ideal behavior, is 37.4 torr.