Question

An automobile antifreeze mixture is made by mixing equal volumes of ethylene glycol (d = 1.114 g/mL; M = 62.07 g/mol) and water (d = 1.00 g/mL) at 20°C. The density of the mixture is 1.070 g/mL. Express the concentration of ethylene glycol as (a) volume percent 50 % v/v (b) mass percent 52.7 % w/w (c) molarity M (d) molality m (e) mole fraction

Answer 1

(a) The volume percent is, 50.63 %

(b) The mass percent is, 52.69 %

(c) Molarity is, 9.087 mole/L

(d) Molality is, 17.947 mole/L

(e) Moles fraction of ethylene glycol is, 0.244

Explanation : Given,

Density of ethylene glycol = 1.114 g/mL

Molar mass of ethylene glycol = 62.07 g/mole

Density of water = 1.00 g/mL

Density of solution or mixture = 1.070 g/mL

According to the question, the mixture is made by mixing equal volumes of ethylene glycol and water.

Suppose the volume of each component in the mixture is, 1 mL

First we have to calculate the mass of ethylene glycol.

`\text{Mass of ethylene glycol}=\text{Density of ethylene glycol}* \text{Volume of ethylene glycol}=1.114g/mL* 1mL=1.114g`

Now we have to calculate the mass of water.

`\text{Mass of water}=\text{Density of water}* \text{Volume of water}=1.00g/mL* 1mL=1.00g`

Now we have to calculate the mass of solution.

Mass of solution = Mass of ethylene glycol + Mass of water

Mass of solution = 1.114 + 1.00 = 2.114 g

Now we have to calculate the volume of solution.

`\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=(2.114g)/(1.070g/mL)=1.975mL`

(a) Now we have to calculate the volume percent.

`\text{Volume percent}=\frac{\text{Volume of ethylene glycol}}{\text{Volume of solution}}* 100=(1mL)/(1.975mL)* 100=50.63\%`

(b) Now we have to calculate the mass percent.

`\text{Mass percent}=\frac{\text{Mass of ethylene glycol}}{\text{Mass of solution}}* 100=(1.114g)/(2.114g)* 100=52.69\%`

(c) Now we have to calculate the molarity.

`\text{Molarity}=\frac{\text{Mass of ethylene glycol}* 1000}{\text{Molar mass of ethylene glycol}* \text{Volume of solution (in mL)}}`

`\text{Molarity}=(1.114g* 1000)/(62.07g/mole* 1.975L)=9.087mole/L`

(d) Now we have to calculate the molality.

`\text{Molality}=\frac{\text{Mass of ethylene glycol}* 1000}{\text{Molar mass of ethylene glycol}* \text{Mass of water (in g)}}`

`\text{Molality}=(1.114g* 1000)/(62.07g/mole* 1kg)=17.947mole/kg`

(e) Now we have to calculate the mole fraction of ethylene glycol.

`\text{Mole fraction of ethylene glycol}=\frac{\text{Moles of ethylene glycol}}{\text{Moles of ethylene glycol}+\text{Moles of water}}`

`\text{Moles of ethylene glycol}=\frac{\text{Mass of ethylene glycol}}{\text{Molar of ethylene glycol}}=(1.114g)/(62.07g/mole)=0.01795mole`

`\text{Moles of water}=\frac{\text{Mass of water}}{\text{Molar of water}}=(1g)/(18g/mole)=0.0555mole`

`\text{Mole fraction of ethylene glycol}=(0.01795mole)/(0.01795mole+0.0555mole)=0.244`