[alumina]

Hi,
I have a problem with an exercise on molar volumes, and I do not know where to start in the resolution! here is the text :

at 25°C, the total volume of ethanol-water mixture agrees with the expression

$$ \frac{V}{cm^3} =1002.93 + 54.666b-0.36394b^2+0.028256b^3 /$$

where b is the numeric value of molality of ethanol. which are the partial molar volumes of water and ethanol for a mixture 0.1 m?

thanks

[enahs]

What does the 0.1 m tell you?

[alumina]

0,1 molale. the concentration expressed in Molality.

[fledarmus]

Deeper - what does it tell you about the relative amounts of water and ethanol in your mixture?

These might also be helpful:

http://en.wikipedia.org/wiki/Partial_molar_volume

http://www.chem.arizona.edu/~salzmanr/480a/480ants/mixpmqis/mixpmqis.html

[alumina]

So, I know that the molality is the ratio of solute's moles and kg's solvent.
In this exercise the water is the solvent. Now,  If I establish that the amount of water is 1 kg , I will have that the solute's moles is 0.1 and its molality is 0.1.

If I substitute the value in the equation , I will get the total volume of mixture.
Is this right?

[fledarmus]

So, I know that the molality is the ratio of solute's moles and kg's solvent.
In this exercise the water is the solvent. Now,  If I establish that the amount of water is 1 kg , I will have that the solute's moles is 0.1 and its molality is 0.1.

If I substitute the value in the equation , I will get the total volume of mixture.
Is this right?

So far so good - now the partial molar volume of ethanol would be the derivative of the total volume with respect to the number of moles of ethanol, and the partial molar volume of water would be the derivative of the total volume with respect to the number of moles of water (per the University of Arizona reference listed above)

[alumina]

thank you! I understand!