Get the molar mass of an unknown liquid in a dissolution - Chemistry Stack Exchange most recent 30 from chemistry.stackexchange.com 2019-09-15T06:03:37Z https://chemistry.stackexchange.com/feeds/question/38534 https://creativecommons.org/licenses/by-sa/4.0/rdf https://chemistry.stackexchange.com/q/38534 1 Get the molar mass of an unknown liquid in a dissolution Voldemort https://chemistry.stackexchange.com/users/16632 2015-10-06T21:47:49Z 2016-05-04T04:57:01Z <blockquote> <p>A dissolution has $10$g of some unknown liquid and also $90$g of water. The freezing point of this dissolution is $-3.33^\text{o}$C. What is the molar mass of the unknown liquid in g/mol?</p> </blockquote> <p>Alright. First of all, the formula to calculate the freezing point of a dissolution is</p> <p>$$T_f - \Delta_f$$</p> <p>Where</p> <p>$$\Delta_f = \text{m} \cdot K_f$$</p> <p>In our case, it would be</p> <p>$$-3.33 = 0 - (\text{m}\cdot 1.86)$$</p> <p>The molality m is</p> <p>$$\frac{\text{moles of unknown liquid}}{\text{kg of water}} = \frac{\text{moles of unknown liquid}}{0.09}$$</p> <p>We have to solve the equation:</p> <p>$$-3.33=-\frac{\text{moles of unknown liquid}}{0.09}\cdot 1.86$$</p> <p>Which yields</p> <p>$$0.16 = \text{moles of unknown liquid}$$</p> <p>However, I need the <strong>molar mass</strong> of this unknown liquid. How do I get it?</p> https://chemistry.stackexchange.com/questions/38534/-/38537#38537 1 Answer by Buttonwood for Get the molar mass of an unknown liquid in a dissolution Buttonwood https://chemistry.stackexchange.com/users/1782 2015-10-06T22:14:52Z 2016-03-05T01:22:31Z <p>As long as you work with <em>sufficently low</em> concentrations, you do not need the molecular mass of your unkown. Indeed, this was (is?) your initial question.</p> <p>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 <a href="https://en.wikipedia.org/wiki/Colligative_properties" rel="nofollow">colligative properties</a>. What you need is to look up the <a href="https://en.wikipedia.org/wiki/Cryoscopic_constant" rel="nofollow">crysoscopic constant</a>, for water (same page) of $1.853 \mathrm{~K·kg/mol}$. </p> <p>On the other side, if you already state the equation of $0.16$ moles corresponding to $10\mathrm{~g}$.</p> <p>Despite advent of mass spectroscopy, such a determination may be still helpful today.</p>