00:01
Hello students, here in this question we have to find out the molarity value that they are provided with the glucose freezing point density and its kf value.
00:09
So here we can use the equation of the depression in freezing point where the delta tf is equal to its constant value into molality.
00:18
So from this we can first find out the molality value.
00:22
So the molality value will be equal to its freezing point minus 11 .8 degree celsius divided by its constant value 1 .86 degree celsius kilogram per mole.
00:34
From this we will get the molality value as 6 .34 mole per kilogram.
00:40
But what we have to find? we have to find out the molarity.
00:44
The molarity is equal to the number of moles divided by the volume of the solvent in liter.
00:50
The density is given.
00:51
So we can convert the kg per liter of the density value by multiplying the density given as 1 .62 gram per ml with 1 kg divided by 1000 gram into 10 raised to 3 ml by 1 liter and we will get the density as 1 .62 kilogram per liter.
01:18
We can now find out the mass of the solvent in 1 liter.
01:22
So the mass of solvent here will be equal to density minus mass of glucose multiplying with its molar mass of the glucose.
01:35
So as a result here we will get it as how much? we will get it as density is 1 .62 but from the molality value we can find out what? easily find out the number of moles of glucose by its molality into 1 kg...