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.

  • $\begingroup$ 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. $\endgroup$
    – Poutnik
    May 2, 2021 at 7:25

1 Answer 1


I am not sure you even need the density in this case.

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: $$\frac{420.62}{420.62+1000 } \approx 29.6\%$$

  • $\begingroup$ You can now upgrade your formatting skills and go beyond using just $\ce{}$, try using $\pu{}$ for physical units. $\endgroup$ May 2, 2021 at 6:21

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