# If 18.6 g of methanol is used to dissolve 2.68 g of Hg(CN)_2, what is the molality of the solution?

$\text{Molality"="Moles of solute"/"Kilograms of solvent} \cong 0.60 \cdot m o l \cdot k {g}^{-} 1$
"Molality"=((2.68*g)/(252.65*g*mol^-1))/(18.6xx10^-3*kg)=??mol*kg^-1??