Question

What is the freezing point of a solution containing 4.78 grams naphthalene (molar mass = 128.2 g/mol) dissolved in 32.0 grams paradichlorobenzene?

Answer 1

This is a freezing point depression problem, so it will use the equation:

ΔT = i Kf m

i = 1 (naphthalene does not dissociate further when dissolved)
Kf = 7.10 C/m (a constant for paradichlorobenzene, which you'd be given)
m = moles of solute / kg of solvent; moles of solute = 4.78 / 128.2 = 0.0373; kg of solvent = 0.032; m = 0.0373 / 0.032 = 1.166m

ΔT = 1 x 7.10 x 1.166 = 8.279 C

This means the normal freezing point of pure paradichlorobenzene is decreased by 8.279 C; the normal freezing point (again, something you'd be given) is 53.5 C, so the new freezing point would be 53.5 - 8.279 = 45.221 C.