Question

1) Benzene freezes at 5.45 degrees celsius. The Kf for benzene is -5.07C/m. What would be the freezing point of a 0.210m solution of octane in benzene?

I am not sure how to start this. I know the equation for freezing point is Kfp * m solute.

Answer 1

So plug in the numbers. That's always a good place to start.
But I caution you that the freezing point is NOT Kfp*m. It's DELTA T that is equal to Kfp*m

Answer 2

would it be delta T = -5.07C/m * 0.210m?

Answer 3

Yes, then the freezing point will be normal f.p. - delta T = new f.p.