# Partial molar volume of binary mixture

A 2% (weight %) solution of a salt with Mr = 190.6 g mol-1 was prepared at 25 0C. The density of the prepared solution was 1.00926 g cm-3. The density of water at 25 0C is 0.99707 g cm-3, and the density of the crystalline salt was 3.00845 g cm-3.

a) ) What is the molar volume of the crystalline salt?

molar volume of salt=(190.6g/mol)/(3.00845g/cm^3)= 63.4cm^3 /mol

b) what's the partial molar volume of the salt at infinite dilution

assuming the mass of solution =100g, w%:(mass of salt)/(mass of solution) x 100

w%: (mass of salt)/100g*100= 2%

mass of salt= 2g

mass of water=100-2=98g

no.of mole salt= 2/190.6=0.010493mol

no.of mole of water= 98/18.02=5.4384mol

x_salt= 0.010493/(5.4384+0.010493) = 0.001925

x_water=1-0.001925=0.99881

these are my workings for part a and b. My understanding of part b is to find partial molar volume via dV/dn, but I can't seem to find the difference in volumes, let alone infinite dilutions...I'm not even sure if I'm on the right track. Do I have any misconceptions?

-struggling physical chem student

• Please, always use as the first step a general algebraic solution, for 4 reasons.: 1/better grasp of principles for OP 2/ better orientation and spotting eventual mistakes 3/solution reusability and permanent value 4/ The site focus is on helping understanding principles and solution procedures, not on solving particular tasks. Use eventually MathJax for mathematical/chemical expression/formula formatting. – Poutnik Oct 27 '20 at 18:09
• Hint: What would be the solution volume without the salt ? – Poutnik Oct 27 '20 at 18:16
• I see. Sorry about that – poorchemstudent Oct 28 '20 at 2:24