Question
An aqueous $\ce{KNO3}$ solution has a molality of $\pu{4.16 m}$ and a density of $\pu{1.08 g/mL}$. Calculate the percentage by mass $\ce{KNO3}$ of the solution.
My Approach
I know that molality ($b$) is defined as:
$$b = \frac{n_\mathrm{solute}}{m_\mathrm{solvent}}$$
In order to the mass of solution I multiplied $\pu{(1.08 g/mL)}\times\pu{1 mL}$, since, $m_\mathrm{solution} = m_\mathrm{solvent} + m_\mathrm{solute}$
However, I don't know where to use this information. I am also not sure what I to do with the $\pu{4.16 m}$ since I don't have moles of solute or mass of solvent. I am thinking that I have to assume some number here, but I might be wrong about this. Any help will be appreciated.