# How many kilograms of solvent are there in a sample of 0.30 molal solution if the sample contains 13 moles of solute?

$43 \text{ Kg of solvent}$
The molality ($m$) is defined as: $m = \frac{n}{\text{Kg of solvent}}$,
where, $n$ is the number of mole of solute.
Therefore, $\text{Kg of solvent"=n/m=(13cancel(mol))/(0.30cancel(mol)/(Kg))=43" Kg of solvent}$