Question

What is the molarity, molality, and mole fraction of ethylene glycol $ \left( {{C_2}{H_6}{O_2}} \right) $ in an aqueous solution that contains $ 40\% $ by mass of the solute? The density of the solution is $ 1.06gm{l^{ - 1}} $

Answer 1

Hint: There are different terms to express the concentration. Molarity, molality, and mole fraction are some of the terms. Molarity is the ratio of number of moles and volume of solution in litres, molality is the ratio of number of moles and mass of solvent in kilograms, mole fraction is the ratio of the number of moles of solute to the moles of solute and moles of solvent.

Complete answer:
Molarity:
Molarity is also known as molar concentration. It can be obtained by dividing the number of moles of solute by volume of solution in litres.
The number of moles of ethylene glycol will be $ \dfrac{{4g}}{{62.07gmo{l^{ - 1}}}} $ which is equal to $ 0.6444mol $
Volume of solution is equal to the ratio of the mass of solution and density $ \dfrac{{100g}}{{1.05gm{l^{ - 1}}}} = 95.24ml = 0.0952L $
The volume of solution in litres will be $ 0.0952L $
Thus, molarity will be $ \dfrac{{0.6444}}{{0.0952}} = 6.76M $
Molality:
Molality can be obtained by dividing the mass of solute with mass of solvent in kilograms
The mass of solvent in kilograms will be $ 0.06kg $
The number of moles of ethylene glycol is $ 0.6444mol $
Thus, molality will be $ \dfrac{{0.6444mol}}{{0.06kg}} = 10.7m $
Mole fraction:
It is the ratio of moles of ethylene glycol and sum of the moles of ethylene glycol and moles of solvent
Moles of ethylene glycol is $ 0.6444mol $
Moles of water will be $ \dfrac{{60g}}{{18.02gmo{l^{ - 1}}}} = 3.329mol $
Thus, mole fraction of ethylene glycol will be $ \dfrac{{0.6444}}{{3.329 + 0.6444}} = 0.162 $

Note:
The number of moles of any solute is the ratio of the weight of the solute and molar mass of solute. The volume of solution must be taken in litres, if it is taken in litres then the volume should be multiplied by a factor of $ 1000 $ as one litre is equal to $ 1000 $ millilitres.