2 gm C6H5COOH dissolved in 25 gm of benzene shows a depression in freezing point equal to 1.62 k.Molar depression constant for benzene is 4.9 k kg \mol.What I sthe percentage association of acid if it forms double molecules(dimmer) solution?

Freezing point depression is discussed here along with some cryoscopic constants including the constant for benzene (they give a value of 5.12).

Freezing-point depression - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Freezing-point_depression)

The freezing point depression is given as ΔTf = Kf · mB (Equation 1) where ΔTf is the change in freezing point, Kf is the cryoscopic constant and m is the molality of the solution. C is the molality of the solution.

The molality of the solution, mB, is often calculated from this equation:

mB = msolute · i. (Equation 2)

where msolute is the molality of the solute and "i" is the number of ions or particles formed in solution. It's not an exact science.

In your case, the freezing point depression is 1.62 Kelvins. You are also given the freezing point molar (not molal for some reason) depression constant (4.9 Kelvin Kilograms / mol).

In this case, you're given 2 grams of benzoic acid. The molecular weight of benzoic acid is 122.12 g/mol. That would give you

\frac {2}{122.12} = 0.0164\,moles of benzoic acid.

However, you are told that benzoic acid forms a dimer in solution. Thus, you have only half that number of moles

\frac {(\frac {2}{122.12})}{2} = 0.00819\,moles of benzoic acid dimer -- IF IT WERE 100% associated (which it's not).

The problem requires that you figure out mB and then figure out "i" from equation 2 (you assume molarities instead of molalities) since your cryoscopic constant is "per mole" instead of "per molal". You'll get a number between 0.0164 moles and 0.00819 moles. You must assume that "i" is linearly dependent on the association.

Give it a try and if you're still having trouble, post your work.