Question

What is definition and formula of 'molal specific volume' ?

Answer 1

It's kind of a misnomer... the "molal specific volume" is really just the molar volume (not molal volume) for the substance:

`barV_i = ((delV_i)/(deln_i))_(T,P,n_j, i ne j)`

i.e. the variation of the volume for substance `i` due to its variation in mols, at constant solution composition (`n_(j ne i)`), temperature (`T`), and pressure (`P`).

For this, we associate the total volume of the solution with the molar volumes and moles of its components:

`V = sum_i n_i barV_i`

NOTE: if you wanted to calculate the total volume of a water-ethanol mixture (which has negative deviation from Raoult's law), you would need to know the molar volumes of both substances at the appropriate temperature and pressure, AND solution composition:

`barV_("EtOH") = "0.05841 L/mol"` at `20^@ "C"` and `"1 bar"`
`barV_("H"_2"O"(l)) = "0.01805 L/mol"` at `20^@ "C"` and `"1 bar"`

IF BY THEMSELVES. `barV` changes with concentration, as the intermolecular forces between solute and solvent increase at higher concentration.

Suppose you mixed `"58.0 mL"` ethanol with `"18.0 mL"` of water. It would not be `"76.0 mL"` of solution, as they interact in solution to reduce the amount of dispersion forces occurring amongst ethanol molecules.

From their pure densities,

`"58.0 mL" xx "0.7893 g"/"mL" xx ("1 mol")/("46.07 g") = "0.9937 mols ethanol"`

`"18.0 mL" xx "0.9982071 g"/"mL" xx ("1 mol")/("18.015 g") = "0.9974 mols water"`

Therefore:

`color(red)V = n_1barV_1 + n_2barV_2`

`= "0.9937 mols EtOH" xx "0.05841 L"/"mol" + "0.9974 mols water" xx "0.01805 L"/"mol"`

`=` `color(red)"76.05 mL"`

And this would appear to be close to the predicted `"76.0 mL"` from additivity, but...

...the actual measured volume is around `"74.3 mL"` instead, if the mixture is `50%` ethanol and `50%` water. That would indeed be in line with our prediction of negative deviation.