# 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.

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.