# Finding mass percent through molality of potassium nitrate solution

### 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.

• Density is excessive input, not needed to calculate conversion. All you need is molar mass of KNO3. Density would be needed for calculation from molarity. May 2, 2021 at 7:25

Say you have $$\ce{1kg}$$ or $$\ce{1000g}$$ water (the solvent). That would mean that there is $$\ce{4.16 mol}$$ of $$\ce{KNO_3}$$.
Since we can figure out the molar mass of $$\ce{KNO_3}$$ = $$\pu{101.11g mol-1}$$, the mass of $$\ce{KNO_3}$$ in the sample must be $$(4.16)(101.11)= \pu{420.62 g}$$.
We already assumed that we had $$\ce{1000g}$$ of water, the mass percent of $$\ce{KNO_3}$$ is then equal to: $$\frac{420.62}{420.62+1000 } \approx 29.6\%$$
• You can now upgrade your formatting skills and go beyond using just $\ce{}$, try using $\pu{}$ for physical units. May 2, 2021 at 6:21