Answer:
Ca(NO3)2 has the highest boiling point ( option A)
Explanation:
Step 1: Data given
A. 1.25 M Ca(NO3)2
B. 1.25 M KNO3
C. 1.25 M CH3OH
D. 2.50 M C6H12O6
Step 2: Calculate highest boiling point
The boiling point depends on the van't Hoff factor
This shows the particles produced when the substance is dissolved. For non-electrolytes dissolved in water, the van' t Hoff factor is 1.
Ca(NO3)2 → Ca^2+ + 2NO3- → Van't Hoff factor = 3
KNO3 → K+ + NO3- → Van't Hoff factor = 2
CH3OH is a non-elektrolyte → Van't Hoff factor = 1
C6H12O6 is a non-elektrolyte → Van't Hoff factor = 1
Ca(NO3)2 has the highest boiling point
The structure is in attachment.
Hexane is alkane (acyclic saturated hydrocarbon, carbon-carbon bonds<span> are </span>single) <span>of six </span>carbon atoms. Methoxy<span> group is the functional group consisting of a methyl group bound to oxygen. Iodo is substituent consists of ionide (element in 17 periodic group).</span>
This problem is providing the heating curve of ethanol showing relevant data such as the initial and final temperature, melting and boiling points, enthalpies of fusion and vaporization and specific heat of solid, liquid and gaseous ethanol, so that the overall heat is required and found to be 1.758 kJ according to:
<h3>Heating curves:</h3>
In chemistry, we widely use heating curves in order to figure out the required heat to take a substance from a temperature to another. This process may involve sensible heat and latent heat, when increasing or decreasing the temperature and changing the phase, respectively.
Thus, since ethanol starts off solid and end up being a vapor, we will find five types of heat, three of them related to the heating-up of ethanol, firstly solid, next liquid and then vapor, and the other two to its fusion and vaporization as shown below:

Hence, we begin by calculating each heat as follows, considering 1 g of ethanol is equivalent to 0.0217 mol:
![Q_1=0.0217mol*111.5\frac{J}{mol*\°C}[(-114.1\°C)-(-200\°C)] *\frac{1kJ}{1000J} =0.208kJ\\ \\ Q_2=0.0217mol*4.9\frac{kJ}{mol} =0.106kJ\\ \\ Q_3=0.0217mol*112.4\frac{J}{mol*\°C}[(78.4\°C)-(-114.1\°C)] *\frac{1kJ}{1000J} =0.470kJ\\ \\ Q_4=0.0217mol*38.6\frac{kJ}{mol} =0.838kJ\\ \\ Q_5=0.0217mol*87.5\frac{J}{mol*\°C}[(150\°C)-(78.4\°C)] *\frac{1kJ}{1000J} =0.136kJ](https://tex.z-dn.net/?f=Q_1%3D0.0217mol%2A111.5%5Cfrac%7BJ%7D%7Bmol%2A%5C%C2%B0C%7D%5B%28-114.1%5C%C2%B0C%29-%28-200%5C%C2%B0C%29%5D%20%2A%5Cfrac%7B1kJ%7D%7B1000J%7D%20%3D0.208kJ%5C%5C%0A%5C%5C%0AQ_2%3D0.0217mol%2A4.9%5Cfrac%7BkJ%7D%7Bmol%7D%20%3D0.106kJ%5C%5C%0A%5C%5C%0AQ_3%3D0.0217mol%2A112.4%5Cfrac%7BJ%7D%7Bmol%2A%5C%C2%B0C%7D%5B%2878.4%5C%C2%B0C%29-%28-114.1%5C%C2%B0C%29%5D%20%2A%5Cfrac%7B1kJ%7D%7B1000J%7D%20%3D0.470kJ%5C%5C%0A%5C%5C%0AQ_4%3D0.0217mol%2A38.6%5Cfrac%7BkJ%7D%7Bmol%7D%20%3D0.838kJ%5C%5C%0A%5C%5C%0AQ_5%3D0.0217mol%2A87.5%5Cfrac%7BJ%7D%7Bmol%2A%5C%C2%B0C%7D%5B%28150%5C%C2%B0C%29-%2878.4%5C%C2%B0C%29%5D%20%2A%5Cfrac%7B1kJ%7D%7B1000J%7D%20%3D0.136kJ)
Finally, we add them up to get the result:

Learn more about heating curves: brainly.com/question/10481356