Vapor pressure lowering I have got this problem in my text book,


  Consider these three
  mercury manometers. One of them
  has 1 mL of water placed on top of the
  mercury, another has 1 mL of a 1 m
  urea solution placed on top of the
  mercury, and the third one has 1 mL of
  a 1 m NaCl solution placed on top of
  the mercury. Identify X, Y, and Z with
  these solutions. 




I got the answer in answer sheet is X = ${H_2}O$ , Y = $NaCl$  and Z = ${(NH_2)}_2{CO}$ . Can any one please explain me? Thanks in advance.
The answer hings on the colligative properties of solutes. In essence solutes lower the vapor pressure of the solvent. NaCl splits into two ions (i.e. two solute "particles") and lowers the vapor pressure of water more than urea for which the whole molecule dissolves as one solute "particle." 

If there was no liquid the Hg level would be higher than any of the three tubes shown since there would be a pure vacuum above the Hg. 

The scale of the relative differences in drawing is obviously greatly exaggerated.