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* m\\\\T_f^o-T_f=K_f*\frac{\text{Mass of ethylene glycol}* 1000}{\text{Molar mass of ethylene glycol}* \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* 1.86^oC/m* (38.6* 1000)/(62.07 g/mol* 175Kg)`

`T_f=-6.61^oC`

Therefore, the freezing point of the solution is,
`-6.61^oC`