If 8 liters of ethylene glycol, C2H4(OH) density = 1.113 g/ml, is placed in an automobile radiator and diluted with 32 liters of water, what is the approximate freezing point of the solution in degrees fahrenheit?
Freezing point depression can be evaluated using the following equation:
ΔTf = Kfmi
where ΔTf - freezing point depression, Kf - freezing point molar constant of water (1.86 °C mol/kg), mi - molality of a solution
Molality of solution is:
mi = msolute / (Mrsolute × msolvent)
where msolute - mass of solute, Mrsolute - molecular weight of solute, msolvent - mass of solvent.
Mass of solute (ethylene glycol) equals:
msolute = dsolute × Vsolute
where dsolute - density of ethylene glycol, Vsolute - volume of ethylene glycol.
THe final equation is:
ΔTf = Kf × dsolute × Vsolute / (Mrsolute × msolvent) = 1.86 °C kg/mol × 1.113 kg/L × 8 L / (0.045 kg/mol × 32 kg) = 11.5 oC.
As a result, the appropriate freezing point of the solution:
Tf = -11.5 °C = 11.3 °F
Answer: 11.3 °F
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