00:02
Here in this problem, given a solution of 68 % nitric acid by mass, we have to find the molality of the solution.
00:12
Now, what does that mean? 68 % nitic acid by mass.
00:17
That means 68 .0 gram hno3 is present per 100 gram of solution.
00:38
Now we'll find out the mass of the solvent.
00:46
Here is solution mass 100 gram and solution is solvent plus solute.
00:52
And here is mass of the salute that is 68 gram hno3.
00:57
So we'll find out mass of solvent.
01:01
Mass of solvent that is mass of solution that is 100 gram minus.
01:10
Mass of solute that is 68 gram and we get it 32 gram now we'll convert this mass of solvent in kilogram 32 times 10 to the power negative 3 kilogram since 1 gram is 10 to the power negative 3 kilogram so we get it 0 .0 3 kilogram so we get the mass of solvents so we get the mass of now we'll find out moles of hno3.
01:45
Moles of hno3, that will be mass of hno3, which is 68 gram.
01:53
So let's write that, divided by the molar mass of nitric acid, that is 63 .01 gram per mole.
02:05
And we cancel gram and gram, and we get 1 .0791...