Question

What are the molarity, molality, and mole fraction of ethylene glycol (`"C"_2"H"_6"O"_2`) in an aqueous solution that contains 40 % by mass of the solute? The density of the solution is `"1.06 g/mL"`.

Answer 1

1. MolaLity = `10.74 m`
2. Molarity = `6.76 M`
3. mole fraction = `0.162`

Explanation:

40% ethylene glycol means

mass of ethylene glycol = `"40 g"`
mass of water (solvent) = `"60 g = 0.060 kg"`
mass of solution = `"100 g"`

1. Molality

Molar mass of ethylene glycol = `"62.07 g/mL"`

no. of moles = `"4 g"/("62.07 g.mol"^-1) = "0.6444 mol"`

Molality = `"no. of moles"/"mass of solvent in kg"`

Molality = `"0.6444 mol"/"0.060 kg" = "10.7 m"`

2. Molarity

`"Molarity" = "no. of moles"/"volume of solution in litres"`

`"Density" = "1.05 g/cm"^3`

`"Volume" = "Mass"/"Density"`

`"Volume" = "100 g"/("1.05 g/mL") = "95.24 mL" = "0.0952 L"`

`"Molarity" ="0.6444 mol"/"0.0952 L" = "6.76 M or 6.76 mol/L"`

3. Mole fraction

`"no. of moles of water" = "60 g"/"18.02 g.mol"^"-1" = "3.329 mol"`

`"Total moles" = "(3.329 + 0.6444) mol" = "3.974 mol"`

`"mole fraction of ethylene glycol" = "0.6444"/"3.974" = 0.162"`

Answer 2

`["C"_2"H"_6"O"_2] = "6.83 M"`

`m_("C"_2"H"_6"O"_2) = "10.741 mol/kg"`

`chi_("C"_2"H"_6"O"_2) = 0.1621`

Read further to see how it was done.

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This is just an exercise in flexing the limits of what you have and calculating various types of concentrations.

The solution is aqueous, so the solvent is water, which is why the density is close to `"1 g/mL"`. Knowing the percent by mass, which is:

`"% w/w" = "mass solute"/"mass solution"xx100%`

we can assume `"1000 g"` solvent for convenience (given that the molality is per `"kg"` of solvent) to get:

`"40% w/w" => 0.40 = "x g solute"/"(1000 + x) g solution"`

Solving for `x`, we can get the mass of the solute:

`0.40(1000 + x) = x`

`=> 400 + 0.40x = x`

`=> 400 = (1 - 0.40)x`

`=> x = 400/(1 - 0.40) = 666.bar(66)` `"g solute"`

Therefore, we can get the mols of solute and mols of solvent:

`color(green)(n_"solute") = (666.bar(66) "g solute")/(2xx12.011 + 6xx1.0079 + 2xx15.99"9 g/mol") = color(green)("10.741 mols ethylene glycol")`

`color(green)(n_"solvent") = ("1000 g solvent")/("18.015 g/mol") = color(green)("55.509 mols water")`

From there, we have all the info we need to calculate the concentrations.

MOLARITY

For the molarity:

`color(blue)(["C"_2"H"_6"O"_2]) = "mols solute"/"L solution"`

`= "10.741 mols solute"/((1000 + 666.bar(66)) cancel"g solution" xx cancel"mL"/(1.06 cancel"g") xx "L"/(1000 cancel"mL"))`

`=` `color(blue)("6.83 M")`

MOLALITY

The molality was made simple because we chose the mass of the solvent to be `"1000 g"`, i.e. `"1 kg"`:

`color(blue)(m_("C"_2"H"_6"O"_2)) = "mols solute"/"kg solvent"`

`= "10.741 mols solute"/"1 kg water"`

`=` `color(blue)("10.741 mol/kg")`

Naturally, we chose `"1000 g"` of water so that we couldn't mess up this calculation as long as we got the mols right (dividing by 1 is easy to get right).

MOLE FRACTION

The mol fraction is:

`color(blue)(chi_("C"_2"H"_6"O"_2)) = n_"solute"/(n_"solute" + n_"solvent")`

`= "10.741 mols solute"/("10.741 mols solute" + "55.509 mols water")`

`=` `color(blue)(0.1621)`