00:01
In this question, we've been given some information relating to ammonium sulphide.
00:06
And we've been asked to complete the given information by calculating the missing information.
00:12
So the molality of ammonium solution, this is 0 .086.
00:19
We're looking at the first part of the question.
00:21
And we've been given the density.
00:23
So what we're going to do is to assume one liter of solution.
00:27
This gives us the mass of solution.
00:30
To be equal to 1 .06 by 10 to the power 3 multiplied by 1 litre which gives us a mass of solution that is equal to 1060 in units of grams.
00:45
What we did here was to use the equation.
00:48
Density is equal to mass over volume.
00:52
So we just met the mass subject of the formula.
00:55
Now that we have the mass of the solution we then move on to determine.
01:01
The number of moles of the solute.
01:04
Number of moles of solute it is going to be equal to the molarity multiplied by the volume of solution which is 0 .886 multiplied by 1 liter which gives us the number of moles of solute that is equal to 0 .886 moles.
01:24
What we did here was to say number of moles is equal to the volume multiplied by the concentration...
