Question

Calculate the freezing point and boiling point of a solution containing 7.70 g of ethylene glycol (C2H6O2) in 89.8 mL of ethanol. Ethanol has a density of 0.789 g/cm3.
A. Calculate the freezing point of a solution.
B. Calculate the boiling point of a solution.

Answer 1

Answer:

(A) -117.48°C (B) 80.5368°C

Explanation:

It is given that density of ethanol is 0.789 $g/cm^3$

volume of ethanol=89.8 mL

mass of ethanol = density of ethanol×volume = 0.789×89.8 = 70.8522 g

molar mass of ethylene glycol = 62 g/mole

therefore moles of ethylene glycol in 7.6 g of it = 7.7/62 = 0.1241 mole

molality = moles of ethylene glycol/mass of ethanol in kg $=\frac{0.1241}{0.708522}=1.75153$

also Kf for ethanol = 1.99 C/m

And Kb  for ethanol = 1.22 C/m

boiling point of ethanol = 78.4 °C

freezing point of ethanol = -114° C

(A) depression in freezing point = molality×Kf = 1.75153×1.99 = 3.4855

thus freezing point of solution = -114-3.4855= -117.48°C

(B) elevation in boiling point = molality×Kb = 1.75153×1.22 = 2.13686

thus boiling  point of solution = 78.4+2.13686=80.5368°C