Question on calculations about solubility How to calculate it ?

A non-volatile solute is dissolved in methanol with a solubility of 20 g/100 mL. Calculate the molar mass of the solute if the vapor pressure of this saturated solution at 20°C is 83 mmHg. Given that the density and vapor pressure of methanol are 0.792 g/mL and 95.7 mmHg at 20°C respectively.

Please help. Thank you.

2 Answers
Mar 2, 2018

You need to use Raoult's Law for this, which states that the vapor pressure of solution is lowered when a solute is added by the mole fraction of the solvent.
First off, lets assume you added 20g of this unknown, and so that means that your methanol is 100mL. 100mL*(0.792g/1mL
=79.2g methanol. Methanol is CH3OH, so its molar mass is about 32g/mole.

Methanol Moles: 79.2g(1mole/32g) = 2.475 moles Methanol.

Raoults Law:
"Psolution" = "Psolvent"*X"solvent", where Xsolvent is the mole fraction of the solvent. Solvent moles/total Moles.

83mmHg = 95.7mmHg*X

X=0.867="moles solvent"/"moles solvent + moles solute"

0.867="2.475moles"/"(2.475 moles + moles solute)"

Moles Solute = 0.3793 moles

Molar mass of solute = grams/mole = 20g/0.3793moles
=52.68 g/mole

Mar 2, 2018

I got "52.88 g/mol".


Since you are given the partial pressure of the solution and of the solvent...

P_A = chi_(A(l))P_A^"*"

where:

  • chi_A is the mol fraction of solvent in the solution phase.
  • P_A is the partial pressure of the solvent IN SOLUTION.
  • P_A^"*" is that of the PURE solvent.

i.e. we have Raoult's law. From this we can first get the mol fraction:

chi_(A(l)) = P_A/P_A^"*" = "83 torr"/"95.7 torr"

= 0.8673

0.8673 = "mols solvent"/("mol solute" + "mols solvent")

Knowing its solubility is "20 g solute"/"100 mL solution", we ASSUME that the density of the solution did NOT change due to adding this non-volatile solute...

"100 mL solution" xx "0.792 g methanol"/"mL pure methanol"

~~ "79.2 g methanol"... more or less.

From this,

"79.2 g MeOH" xx "1 mol"/"32.0416 g MeOH" = "2.47 mols"

Therefore:

0.8673 = 2.47/(x + 2.47)

with x = "mols solute".

x = -2.47 + 2.47/0.8673 "mols"

= "0.3782 mols solute"

So this was a good approximation that the density was approximately of the solution.

The mass of the solute was "20 g" for a "100-mL" solution, so:

color(blue)(M_"solute") = "20.0 g"/"0.3782 mols" = color(blue)("52.88 g/mol")