Explanation:
Let the volume of the solution be 100 ml.
As the volume of glycol = 50 = volume of water
Hence, the number of moles of glycol = [tex]\frac{mass}{molar mass}[/tex]
= [tex]\frac{density \times volume}{molar mass}[/tex]
= [tex]\frac{1.1088 \times 50}{62 g/mol}[/tex]
= 0.894 mol
Hence, number of moles of water = [tex]\frac{50 \times 0.998}{18}[/tex]
= 2.77
As glycol is dissolved in water.
So, the molality = [tex]0.894 \times \frac{1000}{49.92}[/tex]
= 17.9
Therefore, the expected freezing point = [tex]-1.86 \times 17.9[/tex]
= [tex]-33.31^{o}C[/tex]
Thus, we can conclude that the expected freezing point is [tex]-33.31^{o}C[/tex].
