Question

The boiling point of ethanol, CH₃CH₂OH, is 78.500 °C at 1 atmosphere. K_b(ethanol) = 1.22 °C/m

In a laboratory experiment, students synthesized a new compound and found that when 13.27 grams of the compound were dissolved in 287.9 grams of ethanol, the solution began to boil at 78.706 °C. The compound was also found to be nonvolatile and a non-electrolyte.


What is the molecular weight they determined for this compound ?

g/mol

Answer 1

Using the formula for elevation in boiling point:

dT = K_b*m

Putting values:

(78.706-78.5) = 1.22*m

Solving we get:

m = 0.168 molal

Using formula:

m = moles/Mass of solvent in kgs

Putting values:

0.168 = moles/0.2879

So,

moles = 0.168*0.2879 = 0.0483

So,

MW = Mass/moles = 13.27/0.0483 = 274.74 g/mol

Hope this helps !