Percent mass and Molality help please?

Question

A)Consider an aqueous solution of isopropyl alcohol, C3H7OH. If the mole fraction of alcohol is 0.3484, what is the percent by mass of alcohol in the solution

B) Consider an aqueous solution of isopropyl alcohol, C3H7OH. If the mole fraction of alcohol is 0.312, what is the molality of alcohol in the solution?

Answer 1

Part A)

A mole fraction of a particular substance in a solution is calculated as,
moles of substance / total moles in solution

We are told that the mole fraction of the isopropyl alcohol is .3484, meaning that for every 1 mole of total solution, .3484 moles are isopropyl alcohol (the rest being water).

We are asked to calculate the percent mass of isopropyl alcohol in the solution.

Percent mass of a given solute is found as,
mass of solute / total mass of solution

So in order to find the percent mass we first must find the mass of the isopropyl alcohol and the mass of the total solution. We can do this by using what we know about the mole fraction.

To make things easier, lets assume we have 1 mole of solution (actually, it doesnt matter how much we assume we have, but an 1 mole amount makes things easy). If we have 1 mole of solution, that means we have .3484 moles of isopropyl alcohol and .6516 moles of water.
We know (or can calculate ourselves) the molar mass of isopropyl alcohol and of water so from this we can find the mass of the isopropyl alcohol and the mass of the water in the solution.

Mass = moles * molar mass

mass of the isopropyl alcohol = .3484 moles * 60.10 g/mol
mass of isopropyl alcohol = 20.94 grams

mass of water = moles of water * 18.02 g/mol
mass of water = 11.74 grams

The total mass of the solution = mass of isopropyl alcohol + mass of water
total mass = 20.94 grams + 11.74 grams = 32.68 grams

So now we can calculate the % mass of isopropyl alcohol.
% mass of isopropyl alcohol = grams of isopropyl alcohol / total grams of solution
% mass = 20.94 grams / 32.68 grams
% mass = .6408

64.08% isopropyl alcohol by mass



Part 2)

Molality is defined as the number of moles of solute per kilogram of solvent.
Molality = moles of solute / kg solvent

So to find molality we need to find the number of moles of isopropyl alcohol and the total mass (in kg) of solvent.
Again, we can do this by using our knowledge of the mole fraction of the isopropyl alcohol in the solution.

Again, we'll assume we have 1 mole to make things easy, but our assuming any arbitrary amount will not affect the results.

If we have 1 mole of solution, we then have .312 moles of isopropyl alcohol and .688 moles of water.

We now know the number of moles of isopropyl alcohol we have in our example solution. We now need to find the mass of the solvent (water) in kilograms.

From the molar mass of water (18.02 g/mol) we can calculate that there is 12.40 grams of water in the solution.
1000 grams = 1 kilogram
so that is .0124 kg of water

So to find the concentration of isopropyl alcohol in terms of molality,
molality = moles of isopropyl alcohol / kg of water
molality = .312 moles / .0124 kg
molality = 25.17 molality