Answer: [tex]T_{f}[/tex] = -29.96 °C
Explanation: A solution has a lower freezing point compared to the pure solvent, due to the solute lowering the vapor pressure of the solvent.
The "new" freezing point is calculated as:
[tex]\Delta T_{f} = K_{f}.m[/tex]
[tex]K_{f}[/tex] is the molal freezing-point depression constant
m is molality concentration, i.e., moles of solute per kilogram of solvent
Before calculating freezing point, let's find moles of ethylene glycol ([tex]C_{2}H_{6}O_{2}[/tex]) in the solution:
Molar mass [tex]C_{2}H_{6}O_{2}[/tex] = 62.08 g/mol
For 500 g:
[tex]n = \frac{500}{62.08}[/tex]
n = 8.05 moles
The molality concentration for 0.5kg of water:
m = [tex]\frac{8.05}{0.5}[/tex]
m = 16.11
The freezing point will be:
[tex]\Delta T_{f} = K_{f}.m[/tex]
[tex]\Delta T_{f} = -1.86*16.11[/tex]
[tex]\Delta T_{f} = -29.96[/tex]
Freezing point of a solution of Ethylene Glycol and Water is -29.96°C
