An ethylene glycol solution is made using 58.5 g of ethylene glycol (C2H6O2) and diluting to a total volume of 500.0 mL. Calculate the freezing point and boiling point of the solution. (Assume a density of 1.09 g/mL for the solution.)
Given mass of ethylene glycol (C2H6O2)= 58.5 g
Volume = 500.0 mL
Density = 1.09 g/mL
Here we have to calculate the the freezing point and boiling point of the solution.
Moles of C2H6O2 can be calculated as :
(58.5 grams C2H6O2) ( 1 mole C2H6O2 / 62.07 grams) = 0.94 moles of C2H6O2
The mass of water used can be calculated as,
(500.0 ml) (1.09 grams / ml) = 545 grams of mixture
(545 grams of mixture) - (58.5g of ethylene glycol) = 486.5 grams 0.4865kg of water used
Now molality is,
(0.94 moles of C2H6O2) / (0.4865 kilograms of water) = 1.932 molal
Calculation of freezing point of the solution.
Freezing point , Tf= m x Kf
where kf = 1.86 oC/m
Tf = ( 1.86 C / molal) (1.932)
= 3.59 C 3.6oC
The freezing-point depression can be calculated as
where T∘f - the freezing point of the pure solvent (water) = 0 oC
Tf - the freezing point of the solution.
Tf = T∘f - ΔTf = 0 - 3.6 oC = -3.6 oC
Thus the freezing point will be -3.6oC.