Question

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

Answer 1

`"Molality"="Moles of solute"/"Kilograms of solvent"~=0.60*mol*kg^-1`

Explanation:

And here,

`"Molality"=((2.68*g)/(252.65*g*mol^-1))/(18.6xx10^-3*kg)=??mol*kg^-1??`