What is the vapor pressure (in mm Hg) of a solution of 12.5 g of glucose ( C6H12O6) in 75.0 g of methanol (CH3OH) at 27 degrees celcius? The vapor pressure of pure methanol at 27 degrees celcius is 140 mm Hg.
moles glucose = grams/molar mass.
moles methanol = grams/molar mass.
Xmethanol = moles CH3OH/total moles.
Psoln = Xmethanol*Pomethanol.
134.728 mm Hg
To calculate the vapor pressure of the glucose-methanol solution, we need to use Raoult's law. According to Raoult's law, the vapor pressure of a component in a solution is proportional to its mole fraction in the solution.
First, we need to find the mole fraction (X) of methanol. The mole fraction is calculated by dividing the moles of methanol by the total moles of all components in the solution.
To find the moles of methanol, we divide its mass by its molar mass:
moles of methanol = mass of methanol / molar mass of methanol
The molar mass of methanol (CH3OH) is:
12.01 g/mol (C) + 1.01 g/mol (H) + 16.00 g/mol (O) + 1.01 g/mol (H) = 32.04 g/mol
moles of methanol = 75.0 g / 32.04 g/mol
To find the moles of glucose, we divide its mass by its molar mass:
moles of glucose = mass of glucose / molar mass of glucose
The molar mass of glucose (C6H12O6) is:
6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol) = 180.18 g/mol
moles of glucose = 12.5 g / 180.18 g/mol
Next, we calculate the total moles of all components in the solution:
total moles = moles of glucose + moles of methanol
Now that we have the moles of methanol and glucose, we can calculate their mole fractions:
X(methanol) = moles of methanol / total moles
X(glucose) = moles of glucose / total moles
After calculating the mole fractions, we can apply Raoult's law to determine the vapor pressure of the solution:
vapor pressure = X(methanol) * vapor pressure of pure methanol
Now that we have all the necessary information, let's calculate the vapor pressure of the solution:
moles of methanol = 75.0 g / 32.04 g/mol = 2.343 mol
moles of glucose = 12.5 g / 180.18 g/mol = 0.0693 mol
total moles = 2.343 mol + 0.0693 mol = 2.4123 mol
X(methanol) = 2.343 mol / 2.4123 mol = 0.9701
X(glucose) = 0.0693 mol / 2.4123 mol = 0.0299
vapor pressure = 0.9701 * 140 mm Hg = 135.8 mm Hg
Therefore, the vapor pressure of the glucose-methanol solution at 27 degrees Celsius is approximately 135.8 mm Hg.