In this example problem, we convert between units of molarity and mole fraction of a solvent for a non-electrolyte solution.
Our problem reads, "An aqueous solution of 2.55 molar glucose has a density of 1.149 grams per milliliter.
What is the mole fraction of water?"
Our equation for mole fraction of a component of a solution is the moles of that component, in this case water,
divided by the total moles in solution, which would be the moles of water plus the moles of glucose.
We have to be careful because this problem asks for the mole fraction of water, the solvent,
instead of the mole fraction of glucose, the solute.
Since glucose is a non-electrolyte, it does not split apart into ions in solution; it remains as molecules.
Let's begin by looking at the equation for the concentration units we're given in the problem.
Molarity is moles of solute divided by liters of solution.
We're not given a quantity of solution that we're working with, so we can make one assumption regarding quantity.
I like to make an assumption based on the denominator of the concentration units I'm given,
so let's assume we're working with 1 liter of solution.
First let's find moles of glucose.
We can use dimensional analysis.
We assume we're working with 1 liter of solution.
We can convert from liters of solution to moles of solute using the molarity: one liter of solution is 2.55 moles of glucose.
This gives us 2.55 moles of glucose.
Next we'll need to find moles of water.
In order to do this we'll need to use the equation mass of solution equals mass of water plus mass of glucose.
So we'll need the mass of glucose to find the mass of water and ultimately the moles of water.
So for convenience we'll make that calculation here.
2.55 moles of glucose can be converted into grams using its molar mass and dimensional analysis.
We get 459.4 g of glucose.
We can find the mass of water by rearranging our mass of solution equation to solve for water.
We already solved for the massive glucose, so now let's find the mass of solution.
Using dimensional analysis,
we can convert our 1 liter of solution into grams of solution using the density of the solution in the problem and 1 liter is 1000 milliliters.
Completing this calculation, we have 1149 grams of solution in 1 liter of solution.
Now we can plug our mass of solution and mass of glucose into our equation and get our mass of water.
We have 689.6 g of water in 1 liter.
Our next step is to use dimensional analysis and the molar mass of water to convert from grams of water to moles of water.
We have 38.28 moles of water in our 1 liter of solution.
Our last step is to plug these numbers into our mole fraction equation at the top.
Completing the calculation, we should get 0.938 for the mole fraction of water.
In this problem we solved for the mole fraction of the solvent.
This is useful when determining vapor pressure of a solution.
However since glucose is a non-electrolyte, if we had solved for the mole fraction of glucose, the solute,
instead of the mole fraction of water, the solvent, we should have gotten 1 - 0.938 or 0.062 for the mole fraction of glucose.
You can follow the process outlined in this video to see if you can calculate the mole fraction of glucose on your own.