Question

An antifreeze solution is prepared from \[222.6{\text{g}}\] of ethylene glycol \[\left( {{{\text{C}}_2}{{\text{H}}_6}{{\text{O}}_2}} \right)\] and \[{\text{200g}}\] of water. Calculate the molality of the solution. If the density of the solution is \[{\text{1}}{\text{.072g/mL}}\], then what shall be the molarity of the solution?

Answer 1

Hint: Molarity is the number of moles of solute present in one litre of solution. Molality is the number of moles of solute present in \[{\text{one kilogram}}\] of solvent. For a binary solution, the component presence in excess is solvent and the other component present is solute.

Complete step by step answer:
The mass of ethylene glycol is greater than the mass of water. Hence, consider ethylene glycol as a solvent and water as a solute.
The molecular weight of water is \[{\text{18 g/moles}}\].
Calculate the number of moles of water by dividing its mass with molecular weight:

\[\dfrac{{200}}{{18}} = 11.11{\text{ moles}}\]

The number of moles of water present in \[1{\text{kg}}\] of ethylene glycol will be the molality of the solution.
Convert the unit of mass of ethylene glycol from grams to kilograms by dividing with 1000.
\[\dfrac{{{\text{222}}{\text{.6g}}}}{{{\text{1000g/kg}}}}{\text{ = 0}}{\text{.2226kg}}\]

Calculate the molality of solution by dividing the number of moles of water with mass (in kilograms) of ethylene glycol.

\[\dfrac{{{\text{11}}{\text{.11mol}}}}{{{\text{0}}{\text{.2226kg}}}}{\text{ = 49}}{\text{.9m}}\]

Add mass of ethylene glycol to the mass of water to obtain total mass of solution
\[{\text{222}}.{\text{6}} + {\text{2}}00 = {\text{422}}.{\text{6 g}}\]
The density of solution is \[{\text{1}}.0{\text{72 g}}/{\text{mL}}\]
Calculate the volume of solution by dividing the mass of solution with its density
\[{\text{Volume of solution = }}\dfrac{{{\text{Mass}}}}{{{\text{Density}}}}{\text{ = }}\dfrac{{{\text{422}}{\text{.6 g}}}}{{{\text{1}}{\text{.072 g/mL}}}}{\text{ = 394 mL = 0}}{\text{.394 L }}\]

To obtain the molarity of the solution, divide the number of moles of solute with volume (in litres) of solution.
\[{\text{Molarity of solution = }}\dfrac{{{\text{Moles of solute}}}}{{{\text{Volume of solution in L}}}}{\text{ = }}\dfrac{{{\text{11}}{\text{.11 mol}}}}{{{\text{0}}{\text{.394 L}}}}{\text{ = 28}}{\text{.2 M}}\]

Hence, the molarity of the solution is \[{\text{28}}{\text{.2 M}}\].

Note: Molarity and molality are two ways of expressing the concentration of the solution. Other ways of expressing the concentration of the solution are normality, formality, mole fraction, mass percentage, volume percentage, mass by volume percentage and parts per million.