The boiling point of a solution of 6.4 g of vanillin in 50.0g of ethanol is 77.47 degree C.
Change in boiling point of solution
ΔTb = (Kb)(m), where
ΔTb = change in temperature
Kb = elevation constant = 1.22 °C./m
m = molality of the solution
Following the calculation of the moles of vanillin, the molality of vanillin in ethanol will be determined..
Moles of vanillin = 6.4g / 152.12g/mol = 0.042 mol
m of vanillin = 0.042mol / 0.05kg = 0.84m
putting all values on the above equation,
ΔTb = (1.22)(0.84) = 1.0248 °C.
Given;
boiling point of ethanol = 78.5 °C.
So the boiling point of vanillin = 78.5 - 1.0248 = 77.47 °C
Hence, the boiling point of a solution of vanillin is 77.47 °C.
Learn more about the boiling point with the help of the given link:
brainly.com/question/1514229
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