Answer:
The expected freezing point of a 1.75 m solution of ethylene glycol is -3.26°C.
Explanation:
[tex]\Delta T_f=T-T_f[/tex]
[tex]\Delta T_f=K_f\times m[/tex]
[tex]m=\frac{\text{mass of solute}}{\text{Molar mass of solute}\times {Mass of solvent in kg}}[/tex]
where,
[tex]T[/tex] = Freezing point of solvent
[tex]T_f[/tex] = Freezing point of solution
[tex]\Delta T_f[/tex] =depression in freezing point
[tex]K_f[/tex] = freezing point constant
m = molality
we have :
Mass of ethylene glycol = 59.0 g
Molar mass of ethylene glycol = 62.1 g/mol
Mass of solvent i.e. water = 543 g = 0.543 kg ( 1 g = 0.001 kg)
[tex]K_f[/tex] =1.86°C/m ,
[tex]m =\frac{59.0 mol}{62.1 g/mol\times 0.543 kg}=1.75 m[/tex]
[tex]\Delta T_f=1.86^oC/m \times 1.75m[/tex]
[tex]\Delta T_f=3.26^oC[/tex]
Freezing point of pure water = T = 0°C
Freezing point of solution = [tex]T_f[/tex]
[tex]\Delta T_f=T-T_f[/tex]
[tex]T_f=T-\Delta T_f=0^oC-3.26^oC=-3.26^oC[/tex]
The expected freezing point of a 1.75 m solution of ethylene glycol is -3.26°C.
