The comment made by the teacher explains what to do. Look up the freezing and boiling points of pure benzene and apply the appropriate corrections.
You did not calculate the freezing and boiling points. You calculated the changes from the pure compound (benzene), and you did not say if the corrections were up or down.
For PURE benzene,
Tf = 5.51 C
Tb = 80.10 C
You calculated delta T which is in the formula delta T = Kf*m and delta T = Kb*m
The problem asked for freezing point and boiling point.
What is the normal freezing point of benzene. It's approximately 5.50 C (you should confirm that and use the number in your text/note table. Then 5.50-3.39 = ?(I didn't check your 3.39 since I don't have the Kf and Kb memorized for benzene.)
The normal boiling point for benzene is approximately 80 C (again, use the number in your text/notes) so the new boiling point will be 80 + 1.67 (and I didn't confirm the 1.l67 for the same reasons I cited above.).
But this should take care of it I think.
So do I subtract 3.39- 5.51
and 1.67- 80.10? or are those the answers?
I'm still confused.
I got it! Thanks for the help...both of you :)
No. The normal f.p. for benzene is about 5.5. Your delta T is 3.39. That 3.39 is how much the f.p. has been lowered so the new f.p. is 5.l5-3.39 = ?
The 1.67 is delta T for the boiling point. That tells you how much the b.p. has been raised. So the new b.p. is the old one (about 80) + 1.67 = ?
Remember what happens. When a non-volatile solute is added to a volatile solvent three things happen. The freezing point is lowered, the boiling point is raised, and the vapor pressure is lowered. The extent to which these three things happen depends SOLELY upon the number of dissolved particles (;i.e., not the solute or the solvent).
Ohhh yeah i got it already. Thanks.