00:02
Hi there.
00:03
In this question, we have a dilution problem, where we have a solution of a certain concentration, and we are trying to make a new solution that is a lesser concentration.
00:16
So for any dilution problem, we can use the equation m1 times v1 equals m2 times v2.
00:27
So m is molarity, v is the volume, one represents the initial conditions or that stock solution that we're starting with, and twos represent the diluted solution.
00:39
And this equation works because molarity is moles over volume.
00:44
So if we take moles over volume times volume, we get moles.
00:47
However many moles we're taking out of the first solution, that's the same number of moles that are still in that second solution.
00:53
It's just that we're adding more volume, so it's less concentrated.
00:58
Okay, so with this question in mind, or with this equation in mind, let's look at the numbers that we have here.
01:03
We have a 6 .0 molar stock solution...