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Solved: An ethylene glycol solution contains 21.2 g of ethylene glycol (C2H6O2) in 85.4
Chapter 13, Problem 106P(choose chapter or problem)
An ethylene glycol solution contains 21.2 g of ethylene glycol \(\mathrm {(C_2H_6O_2)}\) in 85.4 mL of water. Calculate the freezing point and boiling point of the solution. (Assume a density of 1.00 g/mL for water.)
Questions & Answers
QUESTION:
An ethylene glycol solution contains 21.2 g of ethylene glycol \(\mathrm {(C_2H_6O_2)}\) in 85.4 mL of water. Calculate the freezing point and boiling point of the solution. (Assume a density of 1.00 g/mL for water.)
ANSWER:
Step 1 of 3
Here, we are going to calculate the boiling point and freezing point of the solution .
Calculation of the amount of ethylene glycol in mol
Mass of water = Density x volume = 1.0 g/L x 85.4 L = 85.4 g
We know,
Molar mass of ethylene glycol = 62.07 g/mol
Mass of ethylene glycol = 21.2 g
Amount of glucose in mol = 21.2 g/62.07 g/mol = 0.3415 mol