00:02
Here, given a solution of 0 .150 gram glycerol in 20 gram of water.
00:14
We have to find the boiling point and freezing point of the solution.
00:27
Molecular weight of glycerol is 92 .09 gram per mole.
00:42
And mass of glycerol is 0 .0 .5 .5 .5.
00:50
1 .150 gram.
00:53
Therefore, number of moles of glycerol is the mass of glycerol, which is 0 .150 gram divided by the molecular weight, 92 .09 gram per mole.
01:19
And the value is 0 .0 .00163.
01:33
So, molality we have to find molality is number of moles, which is 0 .00163 moles divided by the mass of the solvent, which is water in kilogram, which is 0 .00163 moles divided by the mass of the solvent, which is water in kilogram, which is 0 .02, so the value for molality is 0 .082m.
02:06
Now we can find out delta t b freezing boiling point elevation...