The freezing point of ethanol, CH3CH2OH, is -117.300 °C at 1 atmosphere. Kf(ethanol) = 1.99 °C/m

In a laboratory experiment, students synthesized a new compound and found that when 12.70 grams of the compound were dissolved in 216.5 grams of ethanol, the solution began to freeze at -117.431 °C. The compound was also found to be nonvolatile and a non-electrolyte.

What is the molecular weight they determined for this compound?

[tex]\Delta T_f=i\times K_f\times m\\\\T_f^o-T_f=i\times T_f\times\frac{\text{Mass of compound}\times 1000}{\text{Molar mass of compound}\times \text{Mass of ethanol}}[/tex]

where,

[tex]\Delta T_f[/tex] = change in freezing point

[tex]T_f^o[/tex] = temperature of pure ethanol = [tex]-117.300^oC[/tex]

[tex]T_f[/tex] = temperature of solution = [tex]-117.431^oC[/tex]

[tex]K_f[/tex] = freezing point constant of ethanol = [tex]1.99^oC/m[/tex]

i = van't hoff factor = 1 (for non-electrolyte)

m = molality

Now put all the given values in this formula, we get

[tex](-117.300)-(-117.431)=1\times 1.99^oC/m\times \frac{12.70g\times 1000}{\text{Molar mass of compound}\times 216.5g}[/tex]

[tex]\text{Molar mass of compound}=891.10g/mol[/tex]

Therefore, the molecular weight of this compound is 891.10 g/mol