What would be the freezing point of a solution that has a molality of 1.324 m which was prepared by dissolving biphenyl (C12H10)
1 answer:
The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.
A solution is prepared by dissolving biphenyl into naphthalene. We can calculate the freezing point depression (ΔT) for naphthalene using the following expression.
where,
- i: van 't Hoff factor (1 for non-electrolytes)
- Kf: cryoscopic constant
- m: molality
The normal freezing point of naphthalene is 80.26 °C. The freezing point of the solution is:
The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.
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= ![\dfrac{[H^{+}] [A^{-}]}{[HA]}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5BH%5E%7B%2B%7D%5D%20%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
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Explanation:
- When an aqueous solution of a certain acid is prepared it is dissociated is as follows-
⇄
Here HA is a protonic acid such as acetic acid,
- The double arrow signifies that it is an equilibrium process, which means the dissociation and recombination of the acid occur simultaneously.
- The acid dissociation constant can be given by -
=
- The reaction is can also be represented by Bronsted and lowry -
⇄
- Then the dissociation constant will be
=
Here, is the dissociation constant of an acid.