Calculate the volume of solution that has a vapour pressure of 25.8 mbar and contains 200.3 g of galactose, C6H12O6, at a temperature of 22.0 °C. The vapour pressure of pure water at 22 °C is 26.40 mbar, the density of water is 1.00 g mL-1, and assume no change in volume occurs upon

dissolution.

Assuming this is from MasteringChemistry, the correct answer is 862 mL.

To calculate the volume of the solution, we need to use Raoult's law, which states that the partial pressure of a solvent above a solution is equal to the vapor pressure of the solvent multiplied by its mole fraction in the solution.

First, we need to determine the mole fraction of galactose in the solution. We can do this by dividing the moles of galactose by the total moles of solute and solvent in the solution.

The number of moles of galactose can be calculated using its molar mass, which is 180.16 g/mol:

moles of galactose = mass of galactose / molar mass
= 200.3 g / 180.16 g/mol
≈ 1.11 mol

Since the volume of the solution is not given, we can assume it to be 1 liter (1000 mL) for simplicity.

The density of water is given as 1.00 g/mL, so the mass of water in the solution would be:

mass of water = total mass of solution - mass of galactose
= 1000 g - 200.3 g
= 799.7 g

The mole fraction of water (solvent) can be calculated as:

moles of water = mass of water / molar mass of water
= 799.7 g / 18.02 g/mol
≈ 44.41 mol

mole fraction of water = moles of water / (moles of water + moles of galactose)
= 44.41 mol / (44.41 mol + 1.11 mol)
≈ 0.9764

Now we can use Raoult's law to find the partial pressure of water in the solution:

partial pressure of water = vapor pressure of water * mole fraction of water
= 26.40 mbar * 0.9764
≈ 25.78 mbar

Since the partial pressure of water in the solution is given as 25.8 mbar, we can now calculate the volume of the solution as follows:

volume of solution = (partial pressure of water / vapor pressure of water) * volume of solvent
= (25.8 mbar / 25.78 mbar) * 1000 mL
≈ 1000 mL

Therefore, the volume of the solution is approximately 1000 mL.

To calculate the volume of the solution, we need to follow a few steps:

Step 1: Calculate the moles of galactose.
First, we need to calculate the moles of galactose, C6H12O6. The molar mass of galactose can be calculated by adding up the atomic masses of carbon, hydrogen, and oxygen:
6*(12.011 g/mol) + 12*(1.008 g/mol) + 6*(16.00 g/mol) = 180.18 g/mol.

To determine the number of moles of galactose, we can use the formula:
moles = mass/molar mass
moles = 200.3 g / 180.18 g/mol
moles ≈ 1.11 mol

Step 2: Calculate the moles of water.
Since no change in volume occurs upon dissolution, we can assume that the volume of the solution is the sum of the volumes of galactose and water.

To calculate the moles of water, we can use the molar mass of water, which is 18.015 g/mol.
moles = mass/molar mass
mass of water = mass of solution - mass of galactose
mass of solution = volume of solution * density of water

We need to convert the vapor pressure values to atmospheres:
vapor pressure = 25.8 mbar * (1 atm / 1013.25 mbar) = 0.025 atm
vapor pressure of pure water = 26.40 mbar * (1 atm / 1013.25 mbar) = 0.026 atm

Now, we can calculate the mass of the solution:
mass of solution = mass of galactose / (vapor pressure / vapor pressure of pure water)
mass of solution = 200.3 g / (0.025 atm / 0.026 atm)
mass of solution ≈ 207 g

Now, we can calculate the moles of water:
moles of water = (mass of solution - mass of galactose) / molar mass of water
moles of water = (207 g - 200.3 g) / 18.015 g/mol
moles of water ≈ 0.37 mol

Step 3: Calculate the volume of the solution.
Using the ideal gas law, we can express the volume of the solution in terms of the moles of water and the temperature:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in liters)
n = moles of water
R = ideal gas constant (0.0821 L*atm/(mol*K))
T = temperature (in Kelvin)

To convert the temperature to Kelvin:
T (in Kelvin) = T (in °C) + 273.15
T (in Kelvin) = 22.0 °C + 273.15 ≈ 295.15 K

Now, rearranging the ideal gas law equation to solve for volume:

V = (nRT) / P
V = (0.37 mol * 0.0821 L*atm/(mol*K) * 295.15 K) / 0.025 atm
V ≈ 9.48 L

So, the volume of the solution that has a vapor pressure of 25.8 mbar and contains 200.3 g of galactose at a temperature of 22.0 °C is approximately 9.48 liters.

mols galactose = 200.3/180 = approx 1.1

Psoln = XH2O*PoH2O
25.8 = XH2O*26.4
Solve for XH2O
1-XH2O = Xgal

Xgal=ngal/(ngal+nH2O)
Solve for nH2O

nH2O = grams/molar mass H2O
Solv for grams. Convert g to volume using density.