00:01
We have to find out the mass percent of barium chloride in the original sample.
00:10
Okay.
00:11
So first we will write down the given values which are provided in the sample.
00:18
So the weight of the sample is given as 2 .5 grams.
00:26
This is the weight of the mixture which contains barium chloride plus sodium chloride.
00:33
Okay, and the mass of the or we can say weight of the precipitate is equal to 2 .123 grams.
00:47
This is the weight of barium sulfate precipitate.
00:53
Okay, after the addition of excess of sulfate.
00:58
Now, if we calculate the number of moles of barium sulfate, it will be equal to the given weight upon the molar mass.
01:14
Okay, the molar mass of barium sulfate is 233 .38 grams per mole and the given weight it is this 2 .123 grams, which is the mass of barium sulfate precipitate.
01:32
So on inserting these values it will be 2 .123 divided by 233 .38.
01:43
It is equal to 0 .0091 moles.
01:55
Okay.
01:55
So these are the moles of barium sulfate...