Question

What is the freezing point of a solution containing 4.134 grams naphthalene (molar mass = 128.2) dissolved in 30.0 grams paradichlorobenzene? The freezing point of pure paradichlorobenzene is 53.0°C and the freezing point depressing constant Kfp is -7.10°C/m.

I'm not sure how to find the freezing point. The only equation I know that might be relevant is the one involving delta T and Kf

Answer 1

Nevermind, I got it.
Find deltaT using the Kfp given, and molality with the grams of solute and paradichlorobenzene given (1.07488 m). Multiply the two to get -7.63167, and then subtract that from the freezing point (53°C) to get 45.37°C.

Answer 2

That's the one you use.
delta T = Kf*m
m = molality = mols/kg solvent
mols naphthalene = grams/molar mass = 4.134 g/128.2 = ?
kg solvent (that's the paradichlorobenzene) = 30 g = 0.030 kg
Use the mols naphthalene and kg of the paradichlorobenze to solve for molality = m
Plug in m and Kf and solve for delta T
Then knowing freezing point of paradichlorobenzene normally is 53.0 C (given in the problem), subtract delta T to get the new freezing point. Your numvwe should be less than 53.0
Post your work if you get stuck.

Answer 3

Same number I obtained but I didn't use all of those digits. You aren't allowed that many. At the most you aren't allowed more than 3 significant figures.