Question

The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 8.50 g of naphthalene (C10H8) in 425 g of benzene (Kf of benzene = 4.90°C/m)?


I worked this problem for someone just a couple of days ago.
delta T = k<sub>f</sub> m
k<sub>f</sub> = 4.90
m = molality = mols/kg. Convert 8.50 g naphthalene to mols and divide by kg of benzene (0.425).
This will give you delta T. Subtract from 5.5 to give you the new freezing point. Check my work. Check my thinking.

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