The vapor pressure of pure water at 25 degrees Celsius is 23.77 mm Hg. What is the vapor pressure, in mm Hg, of water above a solution that is 1.5m glucose, C6H12O6?
It is 1.5m (molality)
How do you calculate the moles of water with only the information given?
To determine the vapor pressure of water above a 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.
The mole fraction (X) is defined as the ratio of the number of moles of a component to the total number of moles in the solution.
In this case, we have a solution that is 1.5m glucose (C6H12O6). This means that the solution contains 1.5 moles of glucose per liter of solution.
To calculate the mole fraction of water, we need to know the total number of moles in the solution. Since the solution is 1.5m glucose, we can calculate the number of moles of glucose using its molarity and the volume.
However, we are missing the volume of the solution given, so we cannot provide an accurate answer without that information.
Is that 1.5m or 1.5M glucose? It makes a difference.
Find moles glucose.
Find mole water.
XH2O = moles water/total moles.
PH2O = XH2O x 23.77
It's 1.5 m which means 1.5 moles glucose in 1 kg solvent.
moles glucose = 1.5
moles in 1000 g H2O = 1000/18.015 = ??
Then XH2O = moles H2O/total moles.