00:02
Question number 51 is a freezing point depression question.
00:07
However, the solute or the solvent is not the common solvent water, but rather in this case it's benzene.
00:14
They give you the molality of hcl, which is the solute in benzene, and the new freezing point of benzene with hcl present, and they want you to determine whether or not hcl is an electrolyte, meaning it separates into its ions in benzene.
00:35
So to answer this question, what we have to do is look up the freezing point constant of benzene, which is 2 .15 degrees celsius per molal, the actual freezing point of benzene, which is 5 .5 degrees celsius, and then use the freezing point depression equation.
00:56
The change in temperature of the freezing point will be equal to the vaunt -hoff factor, which tells you how many ions are present for a particular.
01:05
Species or how many species are present, multiplied by the freezing point constant of benzene and the molality of the solute hcl.
01:15
So rearrangement of this equation allows us to solve for i.
01:19
If i is anything but one, then that means that more solute would be present than the moles of hl added.
01:29
The only way that this can occur is if more moles of solute or present than the hcl added.
01:37
The only way this can occur is if hcl separates into its ions...