Question

Calculate the vapor pressure at 25 C of an aqueous solution that is 5.50% NaCl by mass. (Assume complete dissociation of the solute)
Please explain thoroughly.

Answer 1

This problem will involve finding the mole fraction of the solvent (in this case, water) and then using Raoult's law to find the vapor pressure as compared to normal water to determine the solution's vapor pressure.

The first step to this problem is to find the mole fraction of water. First, find the number of moles of each solution component using a known quantity of solution. Using 100 grams will make using the mass percentage easy.

Since the solution is 5.45% salt by weight, this means 100 grams of solution contains 5.45 grams of salt. Now, convert this to moles using the molar mass of NaCl, 58.44 grams per mole. This gives:

5.45/58.44 = .0933 mol of salt

Now, since we know 5.45 grams of the solution of salt, the other 94.55 grams must be water. Divide the 94.55 grams of water by the molar mass of water, 18.02 grams per mole.

94.55/18.02 = 5.247 mole of water

Now we know how many moles of each are in 100 grams of solution. Add these two quantities together to find the total number of moles:

.0933 + 5.247 = 5.340 moles in solution

Now we can determine the mole fraction of water in the solution. Divide the moles of water by the total moles:

5.247/5.340 = .9826

This is the mole fraction of water. Now we use Raoult's law, which states that the mole fraction of the solvent times the vapor pressure of the pure solvent equals the vapor pressure of the solution. The vapor pressure of pure water at 25 Celsius is 23.76 mmHg. Using Raoult's law, we get:

23.76 * .9826 = 23.34 mmHg

The vapor pressure of this solution will be 23.34 mmHg

Hope this explanation clears it up for you!