Question

What is the molality of a solution prepared by dissolving 19.9 g of KCI in 750.0 mL of water?

Answer 1

`"Molality"` `=` `"moles of solute"/"kilograms of solvent"`

Explanation:

We assume (reasonably) that we have a mass of `750.0*g` of water.

And thus `"molality"` `=`

`(19.9*g)/(74.55*g*mol^-1)xx1/(0.750*kg)` `~=` `0.4*mol*kg^-1`