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

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.