1
$\begingroup$

A dissolution has $10$ g of some unknown liquid and also $90$ g of water. The freezing point of this dissolution is $-3.3\, ^\text{o}$C. What is the molar mass of the unknown liquid in g/mol?

Alright. First of all, the formula to calculate the freezing point of a dissolution is

$$T_f - \Delta_f$$

Where

$$\Delta_f = \text{m} \cdot K_f$$

In our case, it would be

$$-3.33 = 0 - (\text{m}\cdot 1.86)$$

The molality m is

$$\frac{\text{moles of unknown liquid}}{\text{kg of water}} = \frac{\text{moles of unknown liquid}}{0.09}$$

We have to solve the equation:

$$-3.33=-\frac{\text{moles of unknown liquid}}{0.09}\cdot 1.86$$

Which yields

$$0.16 = \text{moles of unknown liquid}$$

However, I need the molar mass of this unknown liquid. How do I get it?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Well you know that you have 0.16 moles of the liquid. What else do you know about the liquid? $\endgroup$ – MaxW Nov 5 '15 at 22:29
  • $\begingroup$ What am I missing? you got that 0,16 mol of the liquid have a mass of 10 g, so its molar mass is 62,5 g/mol $\endgroup$ – fich Feb 14 at 21:55
  • 1
    $\begingroup$ Your liquid could be ethanediol $\ce{C2H4(OH)2}$, with a molar mass equal to $62$ g/mol. $\endgroup$ – Maurice Jun 14 at 20:42
0
$\begingroup$

As long as you work with sufficently low concentrations, you do not need the molecular mass of your unkown. Indeed, this was (is?) your initial question.

Freezeing point depression (salting the footwalk to melt ice in the winter), elevation of the boiling point (adding salt to cooking water that increases it's boiling temperature, too) and osmotic pressure are colligative properties. What you need is to look up the crysoscopic constant, for water (same page) of $1.853 \mathrm{~K·kg/mol}$.

On the other side, if you already state the equation of $0.16$ moles corresponding to $10\mathrm{~g}$.

Despite advent of mass spectroscopy, such a determination may be still helpful today.

Improve this answer
$\endgroup$
1
  • $\begingroup$ So low that it doesn't really change the melting point? Because if it changes it at all, something that weighs 50 g/mol will have twice the effect of something that weighs 100 g/mol, assuming the same mass. $\endgroup$ – SendersReagent Mar 5 '16 at 1:44

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.