Question

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?

Answer 1

Freezing point depression can be evaluated using the following equation:

ΔT_f = K_fm_i

where ΔT_f - freezing point depression, K_f - freezing point molar constant of water (1.86 °C mol/kg), m_i - molality of a solution

Molality of solution is:

m_i = m_solute / (Mr_solute × m_solvent)

where m_solute - mass of solute, Mr_solute - molecular weight of solute, m_solvent - mass of solvent.

Mass of solute (ethylene glycol) equals:

m_solute = d_solute × V_solute

where d_solute - density of ethylene glycol, V_solute - volume of ethylene glycol.

THe final equation is:

ΔT_f = K_f × d_solute × V_solute / (Mr_solute × m_solvent) = 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:

T_f = -11.5 °C = 11.3 °F

Answer: 11.3 °F