1
$\begingroup$

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.

enter image description here

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.

$\endgroup$
  • $\begingroup$ Hint - en.wikipedia.org/wiki/Colligative_properties $\endgroup$ – MaxW Jan 29 '17 at 3:16
  • $\begingroup$ @MaxW but i'm confused whether Y = NaCl or not, i think Z = NaCl is more preferable $\endgroup$ – IAmBlake Jan 29 '17 at 3:32
  • 1
    $\begingroup$ NaCl splits into two ions thus has more colligative influence than urea which the molecule dissolves as one thing. Scale in drawing obviously greatly exaggerated. $\endgroup$ – MaxW Jan 29 '17 at 3:45
  • $\begingroup$ @MaxW sorry i couldn't get it. Can you please write your answer with proper description ? $\endgroup$ – IAmBlake Jan 29 '17 at 3:51
4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Interesting!! The problem statement says urea and the answer does give empirical formula for glucose. I didn't catch that. $\endgroup$ – MaxW Jan 29 '17 at 4:16
  • 1
    $\begingroup$ sorry edited..:) so now Z = urea ? $\endgroup$ – IAmBlake Jan 29 '17 at 4:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.