00:01
In part a, we are to calculate the freezing point depression and boiling point elevation for 5 grams of sodium chloride in 25 grams of water.
00:19
When sodium chloride is placed into water, it is considered an electrolyte and will ionize, which will yield two ions in aqueous solution.
00:34
We'll need that number later.
00:38
We'll then need to calculate the molality of the solution by first finding out the number of moles of nacl.
00:56
0 .855 moles.
01:01
Therefore, the molality is moles solutes divided by mass solvent, moles of solute, which is calculated.
01:22
The mass of the solvent, 25 grams of water, converted into kilograms.
01:35
Dividing will give us a molality of 3 .42 moles per kilogram or just lowercase m for units.
01:49
Change in freezing point for an electrolyte is the formula i constant, freezing point depression constant for water times a malality.
02:01
The i is the degree of ionization.
02:03
Sodium chloride and water produces two ions, n -a -plus and cl -minus, which we saw from the equation earlier.
02:11
The constant for water is 1 .86 times the molality, which will give us 12 .7 degrees celsius.
02:30
The freezing point of water is 0 degrees celsius minus the change 12 .7 degrees celsius for a freezing point.
02:42
Depression of negative 12 .7 degrees celsius...