Question

An aqueous solution of dibasic acid containing 35.4g of the acid per litre of the solution has density 1.0077g/cm3 express the concentration of solution in as many ways you can

Answer 1

Name of the Acid is not given. Neither the Molar mass is given. Although Density of the Solution is given, by which we can assume that the acid is Succinic acid. 
(Question is incomplete in that. There must be molar mass of the dibasic acid or name of the acid.)


Given conditions ⇒
Mass of the acid in its aqueous solution = 35.4 g.
Molar mass of the succinic Acid = 118 g/mole.
Volume of the solution = 1 liter. = 1000 cm³.


Now, Let us First expressed the concentration of solution in Molarity.

∵ Molarity = No. of moles of the solute (or Acid)/Volume of the Solution in liter.

No. of moles of the solute or acid = Mass/Molar Mass
= 35.4/118
= 0.3 

∴ Molarity of the solution = 0.3/1
   = 0.3 M.

Hence, the Molarity of the solution is 0.3.


Now, We can also Express the concentration of the solution in terms of Molality.

Given ⇒
Density of the solution = 1.0077 g/cm³.
∵ Density = Mass/Volume
∴ Mass = Density × Volume
∴ Mass = 1.0077 × 1000
∴ Mass of the solution = 1007.7 grams.

We know, Mass of the Solution = Mass of the Solute (or Acid) + Mass of the Solvent (or Water)
∴ Mass of the Solvent (or Water) =  Mass of the Solution - Mass of the Acid.
 = 1007.7 - 35.4
 = 972.3 g
 = 0.9723 kg.

∵ Molality = No. of moles of the Solute (or Acids)/Mass of the solvent in kg.
   = 0.3/0.9723
   = 0.308 m.

Hence, the Molality of the solution is 0.308 m.


Hope it helps.

Answer 2

Answer:

 M=118×100035.4×1000=0.3

 m=Mw2×(Vsol×dsol−W2)W2×1000

=118(1000×1.0077−35.4)35.4×1000

=118×972.335.4×1000=0.31

 N=n×M=2×0.3=0.6N

Weight of solvent (W1) = Weight of solution -Weight of solute

=Vsol×dsol−W2 =1000×1.0077−35.4=972.3

The mole fraction of solute χ2=n1+n2n=Mw1W1