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Solved: An automobile antifreeze mixture is made by mixing
Chapter 14, Problem 58P(choose chapter or problem)
An automobile antifreeze mixture is made by mixing equal volumes of ethylene glycol \((d=1.114 \mathrm{~g} / \mathrm{mL} ; M=62.07 \mathrm{~g} / \mathrm{mol})\) and water \((d=1.00\mathrm{\ g}/\mathrm{mL})\) at \(20^{\circ} \mathrm{C}\) . The density of the mixture is \(1.070\mathrm{\ g}/\mathrm{mL}\) . Express the concentration of ethylene glycol as:
(a) Volume percent
(b) Mass percent
(c) Molarity
(d) Molality
(e) Mole fraction
Questions & Answers
QUESTION:
An automobile antifreeze mixture is made by mixing equal volumes of ethylene glycol \((d=1.114 \mathrm{~g} / \mathrm{mL} ; M=62.07 \mathrm{~g} / \mathrm{mol})\) and water \((d=1.00\mathrm{\ g}/\mathrm{mL})\) at \(20^{\circ} \mathrm{C}\) . The density of the mixture is \(1.070\mathrm{\ g}/\mathrm{mL}\) . Express the concentration of ethylene glycol as:
(a) Volume percent
(b) Mass percent
(c) Molarity
(d) Molality
(e) Mole fraction
ANSWER:
Step 1 of 5
Here, we are going to calculate the concentration of ethylene glycol in different concentration unit.
(a)
Here, we have to determine the concentration of ethylene glycol in volume percent.
We know that,
---(1)
Given that,
Density of ethylene glycol =1.114 g/mL
Density of water (solvent) =1.0 g/mL
Let us consider, the volume of the mixture is 1.0 mL
Therefore, mass of the solvent = 1.0 mL 1.0 g/mL =1.0 g
Then, mass of the solute (ethylene glycol) in 1.0 mL = 1.0 mL 1.114 g/mL
= 1.114 g
Thus, the mass of the solution = (mass of water + mass of ethylene glycol)
=(1.0 g + 1.114 g) = 2.114 g
Therefore, volume of the solution = Mass of solution/Density of solution
= = 1.976 mL
Thus, the volume % of ethylene glycol is given by the given equation:
=
= 50.61 %
Thus, the concentration of ethylene glycol in volume percent is 50.61%.