Solution Found!
Why does the antifreeze ingredient ethylene glycol
Chapter 11, Problem 45P(choose chapter or problem)
Why does the antifreeze ingredient ethylene glycol (\(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\); \(\mathcal{M}=62.07 \mathrm{~g} / \mathrm{mol}\)) have a boiling point of \(197.6^{\circ} \mathrm{C}\), whereas propanol (\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} ; 60.09 \mathrm{~g} / \mathrm{mol}\)), a compound with a similar molar mass, has a boiling point of only \(97.4^{\circ} \mathrm{C}\)?
Equation Transcription:
HOCH2CH2OH
ℳ = 62.07 g/mol
197.6°C
CH3CH2CH2OH; 60.09 g/mol
97.4°C
Text Transcription:
HOCH_2CH_2OH
ℳ = 62.07 g/mol
197.6°C
CH_3CH_2CH_2OH; 60.09 g/mol
97.4°C
Questions & Answers
QUESTION:
Why does the antifreeze ingredient ethylene glycol (\(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\); \(\mathcal{M}=62.07 \mathrm{~g} / \mathrm{mol}\)) have a boiling point of \(197.6^{\circ} \mathrm{C}\), whereas propanol (\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} ; 60.09 \mathrm{~g} / \mathrm{mol}\)), a compound with a similar molar mass, has a boiling point of only \(97.4^{\circ} \mathrm{C}\)?
Equation Transcription:
HOCH2CH2OH
ℳ = 62.07 g/mol
197.6°C
CH3CH2CH2OH; 60.09 g/mol
97.4°C
Text Transcription:
HOCH_2CH_2OH
ℳ = 62.07 g/mol
197.6°C
CH_3CH_2CH_2OH; 60.09 g/mol
97.4°C
ANSWER:
Solution 45P
Here we have to explain why does the antifreeze ingredient ethylene glycol (HOCH2CH2OH; = 62.07 g/mol) have a boiling point of 197.6°C, whereas propanol (CH3CH2CH2OH;