# Finding molar mass of naphthalene using the freezing point of a solution?

Naphthalene is the main ingredient in some mothballs. The freezing point of a solution made by dissolving $\pu{7.01 g}$ of naphthalene in $\pu{200 g}$ of benzene is $4.20 \pu{^\circ C}$ . What is the molar mass of naphthalene?

I started by using $T_f= K_f\times m$ (we're given the $K_f$ of benzene is $5.12 \space{} \pu{ ^\circ C/m}$).

$4.20~ \pu{^\circ C}=(5.12 \pu{^\circ C/m})(\pu{m})$

$m=0.8203125$ molal.

molality = moles of solute per kg of solution

$\implies 0.8203125$ molal= (unknown moles)/(kg solvent)

Am I supposed to use kg of total solvent or kg of naphthalene? I can't seem to figure out where to go next.

## 1 Answer

molality = moles of solute per kg of solution

This is incorrect. Molality instead is moles of solute per kg of solvent. In your case, do you know which one of nepthalene or benzene is the solvent? Whichever it is, use its mass in the formula's denominator.

Moreover, your usage of the formula, along the line "I started by using", is incorrect. The LHS should be $\Delta T_\mathrm f~(=T_\mathrm {f2}-T_\mathrm {f1})$ and not $T_\mathrm f$. The question is incomplete because it should have also mentioned the initial freezing point of pure benzene solvent.