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