mol. (refer to the table for the values of Kb and Kf.). ...

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September 2022 1 2 Report

Calculate the boiling point and freezing point of a solution containing 478 g of ethylene glycol in 3202 g of water. Molar mass of ethylene glycol is 62.01 g/mol. (refer to the table for the values of Kb and Kf.).

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MindOverMatter

Hey!

Boiling point of pure water = 100°C

$\Delta T_{b}=k_{b}\times m_{s}$

$k_{b}=0.51 \: C \: kg/mol$

n(ethylene glycol) $= \frac{478}{62.1} = 7.71 \: mol$

Molality(ethylene glycol) $=7.71 \: mol \times 3.202 \: kg = 2.41 \: mol/kg$

$\Delta T_{b} = 0.51 \times 2.41= 1.23°C$

Boiling point of solution = 100 + 1.23 = 101.23°C

$\Delta T_{f} =$ freezing point of a solvent - freezing point of a solution

$\Delta T_{f}= T_{f0}- T_{f}$

$\Delta T_{f}=k_{f} \times M_{s}$

$k_{f} = 1.86 \: C \: kg/mol$

$\Delta T_{f}= 1.86 \times 2.41 = 4.48°C$

$\Delta T_{f}= 0.0-4.48=\boxed{-4.48°C}$

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