# 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.

• Now that's a nice one. The problem is simple, but the answer may seem so counterintuitive that people would refuse to embrace it. Also, it is not clear how can we help in solving this question, short of telling the answer straight away. OK, let's try: what would be the vapor pressure inside the container? Mind you, I'm not asking about the exact figures (for which there is no data anyway), but for qualitative description, like "slightly higher than something" or "slightly lower than something". Oct 6 '16 at 11:36
• What do you mean by sugar? Cellulose is a sugar like maltose but their structure are really different. Oct 6 '16 at 12:08
• I would assume this is a solution of sucrose. Any soluble sugar would probably give the same result.
– Zhe
Oct 6 '16 at 14:21

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 $\chi_s <1$, then $p < p_w$. How do you use this information?