00:01
So in this problem, first of all, this original solution is a density of 1 .19 grams for 1 milliliter, and it is 37 .0 % hcl.
00:13
Therefore, we are going to multiply that by the density by 0 .370, and then we get the mass of hcl in 1 millimeter of a solution.
00:24
Now we're doing this solution.
00:26
So let's take that density, and then we're going to multiply by how much we, how much of the volume that.
00:35
We are using so we have taken 4 .50 milliliters so multiplied by 4 .50 militers, milliliters cancels out and then we are going to dilute it to 100 milliliters so divide by 100 milliliters because that's the new volume so we took the mass in 4 .5 milliliters and diluted it in 100 militers then we took 10 milliliters of this dilute solution.
01:09
So now multiplied by 10 millimeters.
01:10
So you see milliliters cancels and we're left with the grams of hcl because we're just dealing with the mass within these volumes.
01:18
So now this reacts with the silver nitrate solution.
01:21
You want to know how many millilators of the similarity of silver nitrate is required to precipitate all of the chloride.
01:27
So now like we discussed earlier, we have the mass of the hcl after we perform all of these steps.
01:38
So now convert this mass to the moles of hcl because we need to convert between moles of reactant and the other reactant...