Answer : The freezing point of the solution is, [tex]-6.61^oC[/tex]
Explanation : Given,
Mass of ethylene glycol = 38.6 g
Mass of water = 175 g
Molar mass of ethylene glycol = 62.07 g/mole
Formula used :
[tex]\Delta T_f=K_f\times m\\\\T_f^o-T_f=K_f\times\frac{\text{Mass of ethylene glycol}\times 1000}{\text{Molar mass of ethylene glycol}\times \text{Mass of water}}[/tex]
where,
[tex]\Delta T_f[/tex] = change in freezing point
[tex]T_f^o[/tex] = temperature of pure water = [tex]0^oC[/tex]
[tex]T_f[/tex] = temperature of solution = ?
i = Van't Hoff factor for non-electrolyte solution = 1
[tex]K_f[/tex] = freezing point constant = [tex]1.86^oC/m[/tex]
m = molality
Now put all the given values in this formula, we get
[tex]0^oC-T_f=1\times 1.86^oC/m\times \frac{38.6\times 1000}{62.07 g/mol\times 175Kg}[/tex]
[tex]T_f=-6.61^oC[/tex]
Therefore, the freezing point of the solution is, [tex]-6.61^oC[/tex]
