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.
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
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:
83mmHg = 95.7mmHg*X
X=0.867=
0.867=
Moles Solute = 0.3793 moles
Molar mass of solute = grams/mole = 20g/0.3793moles
=52.68 g/mole
I got
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
"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
color(blue)(M_"solute") = "20.0 g"/"0.3782 mols" = color(blue)("52.88 g/mol")