Question

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.

Answer 1

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

Answer 2

I got `"52.88 g/mol"`.

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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")`