# 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?

##### 1 Answer

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 **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:

#\mathbf(V_"tot" = n_1barV_1 + n_2barV_2)# where:

#V_"tot"# is the total volume of the solution.#n_1# and#n_2# are the#"mol"# s of compound or substance#1# and#2# , respectively.#barV# is the molar volume, whose units are#"L/mol"# . It happens to be the reciprocal of the molar density#barrho# , whose units are#"mol/L"# . Ironically that is a general way of saying the molar concentration, since#"M"# #=# #"mol/L"# .

If we let

#"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:

#"L" = "g solid"xx"L solid"/"g solid" + "g Toluene"xx"L solid"/"g Toluene"#

#color(green)(V_"tot" = m_"solid"/rho_"solid" + m_"Toluene"/rho_"Toluene")# where

#rho# is in#"g/L"# and#m# is in#"g"# , while#V_"tot"# is still in#"L"# .

Now, let's convert the density of toluene to

#"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_"Toluene" ~~ 46.23 - 24.9"9 g" = "21.24 g Toluene"#

Finally, we can get the density of the solid by solving for

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

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

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

The density is therefore:

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

#= ("24.99 g solid " xx " 864" "g/L Toluene")/("0.0500 L total" xx ("864" "g Toluene")/"L" - "21.24 g Toluene")#

#=# #"983.21 g/L"#

#=# #color(blue)("0.9832 g/mL")#