Change in volume of two beakers on reaching equilibrium Two beakers are placed side by side in a sealed container. The first beaker contains 100 ml of pure water. The other an equal volume of an aqueous solution of sugar. What should be the volume of the liquid in the two beakers when all physical change stop, that is when the equilibrium is reached?


I know that the vapour pressure of pure water is higher than the vapour pressure of the sugar solution. But is it anyway going to help me in solving this question.
Okay, so you know that the vapor pressure of the pure water is some value, $p_w$. And you know from Raoult's Law that a solution with a nonvolatile solute of mole fraction $\chi_s$ will have a vapor pressure $p=p_w \chi_w$, where $\chi_w = 1- \chi_s$, so $p=p_w(1-\chi_s)$.

Since $\chi_s &lt;1$, then $p &lt; p_w$. How do you use this information?

Well, as the system approaches equilibrium, the two beakers will want to exchange water through the vapor phase to equilibrate the vapor pressures. What will that look like when it happens?  Can it happen?

By this logic, you can determine where the water is in your system. However, you cannot determine the volume(s) in the beakers without the following additional information:


Volume of the enclosing vessel - helps determine what mass of water remains in the gas phase
Temperature of the system - determines initial vapor pressure of pure water
Concentration of the sugar solution. 
The non-ideal behavior of the sugar-water solution's volume.