The boiling point of ethanol (C₂H₅OH) is 78.5°C. What is the boiling point of a solution of 6.4 g of vanillin M = 152.14 g/mol)

Question

2 years ago

8

in 50.0 g of ethanol (Kb of ethanol = 1.22°C/m)?

1 answer:

Gwar [14] · Answer 1 · 2023-09-06T13:47:42+03:00

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:

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