Question

A solution is made by dissolving 0.656 mol of nonelectrolyte solute in 825 g of benzene. Calculate the freezing point, Tf, and boiling point, Tb, of the solution. Constants may be found here.

Answer 1

You are not providing constants. I will provide equations for you to solve this exercise.

As those values change due the concentration of the solute in bencene. The properties of bencene are going to change. Such as boiling point and freezing point

NEW FREEZING POINT

ΔT = K_f · m where K_f is the constat and m is the molality of the solution (m= 0.8)

now if you solve for deltaT, you will now this is the difference between the pure component freezing point and the new freezing point.

NEW BOILING POINT

ΔT_b = K_b · m

here you have deltaT as the difference between the new boiling point and the pure component boiling point. Solve for the new boiling point.