[SapereAude1490]

Hello. My next (and this year's first) experiment will be to determine the apparent molar volume of NaCl.
And as I started preparing for it, I ran across a problem.

The empirical equation that describes the behaviour of a diluted salt solution is



where V_B is the apparent partial molar volume of component B
         m_B is the molality of that component in mol(B)/kg
         V_B⁰ is the apparent partial molar volume of component B for an infinitely diluted solution
         a is a constant

We can also calculate the apparent volume by



where V is the volume of the solution
         V_A is the volume of solvent
         n_B is the amount of substance B (solute)

According to my handbook, the numerical value of n_B is equal to the molality, because molality is defined as the amount of substance B divided by the mass (1kg) of the solvent. So, the book says, n_B and m_B are (somehow) interchangeable and the the volumes V and V_A are expressed as

and

replacing these two in the second equation I should get



Now, my question is: Can you really just substitute the amount of B (n_B) with molality (m_B)?

And in , where did the m_B come from? Because, if i multiply molality(m_B) with molar mass(M_B), don't I get just a number, with no unit?  - ([mol/kg] * [kg/mol] = 0 )

[SABRY]

1. In this case n_b = m_b since the basis of your calculation is 1 kg of solvent.

2. By definition of molality, in 1 kg of solvent contains m_b mol of solute. This is also designated as n_b. Therefore m_b = n_b.

3. If, for example, your basis of calculation is 10 kg of solvent then, the total number of mole of solute B is:

n_b = 10 M_b

4. When you multiply m_bM_b, the unit is kg.

5. In your calculation, be careful with units consistency.