?chemistry question, please see attached for question Consider the two beakers u

Question

?chemistry question, please see attached for question

Consider the two beakers under a sealed dome. One beaker contains pure solvent, while the other contains a solution of this solvent. If the system contained only a beaker of pure solvent, how would the solvent attain equilibrium with its vapor? If the system contained only the solution, how would the solvent attain equilibrium with its vapor? How does the amount of vapor required to reach equilibrium in the pure solvent compare with that of the solution? Account for the net movement of solvent which has occurred in the "After" picture. It may help to envision equilibrium as a "see-saw" which, when one side becomes too heavy, will tip to the other side to reach a status quo.

Explanation / Answer

(A)When only solvent is there,

Molecules present at the surface leaves as vapours due to unequal forces around itself compared to the molecule in the bulk. Since the solvent beaker is in closed system, the vapours after the collision with walls or among themselves or with surface of solvent, get back to the solvent phase. So, that after sometime it reaches to equilibrium whre rate of evaporation is equal to rate of condensation.

(B) When only solution is present, (a solution of non volatile soute is assumed)

Same as above, solvent molecules of solution will evaporate from surface and get back to solutions state, and finally reaches to equilibrium.

(C) Compared to pure solvent, equilibrium vapoure pressure of solution is smaller. Because, some of the solvent molecules from surface are replaced by non volatile solute molecules and thus less number of molecules find to enter into vapour phase.

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