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Solved: An ethylene glycol solution is made using 58.5 g of ethylene glycol (C2H6O2) and
Chapter 13, Problem 121P(choose chapter or problem)
An ethylene glycol solution is made using 58.5 g of ethylene glycol \(\mathrm{(C_2H_6O_2)}\) and diluting to a total volume of 500.0 mL. Calculate the freezing point and boiling point of the solution. (Assume a density of 1.09 g/mL for the solution.)
Equation Transcription:
Text Transcription:
(C_2H_6O_2)
Questions & Answers
QUESTION:
An ethylene glycol solution is made using 58.5 g of ethylene glycol \(\mathrm{(C_2H_6O_2)}\) and diluting to a total volume of 500.0 mL. Calculate the freezing point and boiling point of the solution. (Assume a density of 1.09 g/mL for the solution.)
Equation Transcription:
Text Transcription:
(C_2H_6O_2)
ANSWER:Solution:
Step 1
Given:
Given mass of ethylene glycol (C2H6O2)= 58.5 g
Volume = 500.0 mL
Density = 1.09 g/mL
Know:
Here we have to calculate the the freezing point and boiling point of the solution.
Explanation:
Moles of C2H6O2 can be calculated as :
(58.5 grams C2H6O2) ( 1 mole C2H6O2 / 62.07 grams) = 0.94 moles of C2H6O2
The mass of water used can be calculated as,
(500.0 ml) (1.09 grams / ml) = 545 grams of mixture
(545 grams of mixture) - (58.5g of ethylene glycol) = 486.5 grams 0.4865kg of water used
Now molality is,
(0.94 moles of C2H6O2) / (0.4865 kilograms of water) = 1.932 molal