Question

Ethylene glycol, C2H6O2, is a nonvolatile substance unable to form ions in water. If 38.6 grams of ethylene glycol is dissolved in 175 grams of water, what is the freezing point of the solution? Kf = 1.86°C/m; Kb = 0.512°C/m
-6.61°C

1.82°C

6.61°C

-1.82°C

Answer 1

Take 36 / (24 + 6 + 32) = 0.580 mol
0.580/.175 = 3.31 m
depression = 1.86 * 3.31 = 6.16
Subtract from 0 gives -6.16 C closest is A

Answer 2

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, $-6.61^oC$