Hey Guys! So today in chem class we went over notes on molarity and molality dilution.

The first concept that we talked about was the concentration of solute. That is, the amount of solute in a solution is given by its concentration. There are four units if concentration.

The first is molartiy and you can compute it by this equation, molarity (M) = moles of solute/liters of solution. This equation comes in hand when you are trying to figure out the mass of a substance needed to make a certain amount of solution at a specific molarity, or to simply find the molarity of a solution.

The second unit in with you can compute the concentration of a solution is molality, be sure to not get this and molarity confused for the sound very similar. With this unit the equation is molality (m) of solution moles of solute/kilograms of solvent.

The third unit in with you can compute the concentration of a solution is percent by mass. The equation of this unit is % by mass = (grams of solute/grams of solution) X 100.

The fourth and final way of compute the concentration of a solution is by finding the mole fraction. To find this you use the equation mole fraction (x) = moles of solute/ moles of solvent + moles of solution. For this you want to remember to have your final answer be in decimals.

The second concept we discussed was molarity and dilution. This concept talked about diluting a solution and this is simply adding more solvent, in our case water, to the solution. This will dilute the solution but remember that the amount of solute does not change and because of this, it is really easy to calculate the new molarity. All you do is take the information you have (at least three points) and plug it into the equation, MaVa=MbVb.

That was all for today’s class. Homework for Monday is to finish your lab and do the webassigns also remember to keep up with your work sheets.

The next scribe is …………………….. Rachel M.