# If the total volume of the resultant "46.23 g" solution that forms after mixing "24.99 g" of a certain solid into a certain volume of toluene (rho_"Tol" = "0.864 g/mL") is "50.0 mL", what is the density of this solid?

Jan 30, 2016

An approach that is something you may have been taught is to assume that the volume is entirely additive (even though it really isn't). Normally, however, that's a pretty good assumption.

Also, since the total volume AFTER adding the solid was given as $\text{50.0 mL}$, there is no doubt that that is the true volume.

If it went unstated when the volume was measured, then that wouldn't be the true volume because of the displacement by the solid causing the liquid to rise to a higher volume marking.

One useful equation (Physical Chemistry: A Molecular Approach, Ch. 24-1) for this situation is:

$\setminus m a t h b f \left({V}_{\text{tot}} = {n}_{1} {\overline{V}}_{1} + {n}_{2} {\overline{V}}_{2}\right)$

where:

• ${V}_{\text{tot}}$ is the total volume of the solution.
• ${n}_{1}$ and ${n}_{2}$ are the $\text{mol}$s of compound or substance $1$ and $2$, respectively.
• $\overline{V}$ is the molar volume, whose units are $\text{L/mol}$. It happens to be the reciprocal of the molar density $\overline{\rho}$, whose units are $\text{mol/L}$. Ironically that is a general way of saying the molar concentration, since $\text{M}$ $=$ $\text{mol/L}$.

If we let ${\overline{V}}_{1}$ belong to the solid, then we can work with this. Since the units in the equation work out as:

$\text{L" = "mols solid"xx"L solid"/"mol solid" + "mols Toluene"xx"L Toluene"/"mol Toluene}$

...under that volume assumption, we can approximate a new equation using the reciprocal mass densities instead of the reciprocal molar densities and write:

$\text{L" = "g solid"xx"L solid"/"g solid" + "g Toluene"xx"L solid"/"g Toluene}$

$\textcolor{g r e e n}{{V}_{\text{tot" = m_"solid"/rho_"solid" + m_"Toluene"/rho_"Toluene}}}$

where $\rho$ is in $\text{g/L}$ and $m$ is in $\text{g}$, while ${V}_{\text{tot}}$ is still in $\text{L}$.

Now, let's convert the density of toluene to $\text{g/L}$ so the units will cancel:

$\text{0.864 g"/cancel"mL" xx ("1000" cancel("mL"))/"1 L" = "864 g/L}$

Next, we can get the approximate mass of toluene to be:

${m}_{\text{Toluene" ~~ 46.23 - 24.9"9 g" = "21.24 g Toluene}}$

Finally, we can get the density of the solid by solving for ${\rho}_{\text{solid}}$:

V_"tot" - (m_"Toluene")/(rho_"Toluene") = (m_"solid")/(rho_"solid")

$\left({V}_{\text{tot"rho_"Toluene" - m_"Toluene")/(rho_"Toluene") = (m_"solid")/(rho_"solid}}\right)$

$\left({\rho}_{\text{Toluene")/(V_"tot"rho_"Toluene" - m_"Toluene") = (rho_"solid")/(m_"solid}}\right)$

The density is therefore:

$\textcolor{b l u e}{{\rho}_{\text{solid") = (m_"solid"rho_"Toluene")/(V_"tot"rho_"Toluene" - m_"Toluene}}}$

$= \left(\text{24.99 g solid " xx " 864" "g/L Toluene")/("0.0500 L total" xx ("864" "g Toluene")/"L" - "21.24 g Toluene}\right)$

$=$ $\text{983.21 g/L}$

$=$ $\textcolor{b l u e}{\text{0.9832 g/mL}}$