Question

Short answer

The 1 mole of binary compound depresses the temperature of 1 kg of water almost twice the maximum amount as much as 1 mole of glycerine thanks to the difference in the temperature depression of electrolyte and non-electrolyte.

Step-by-step solution

Temperature Depression

The temperature depression of electrolytes and non-electrolyte is different.

The greater the concentration, the greater the temperature is going to be depressed.

Explanation in terms of Van’t Hoff factor

This is important due to the equation we use to work out freezing point depression

$\mathbf{\Delta }\mathbf{T}\mathbf{f}\mathbf{r}\mathbf{e}\mathbf{e}\mathbf{z}\mathbf{i}\mathbf{n}\mathbf{g}\mathbf{=}\mathbf{i}\times \mathbf{K}\mathbf{f}\times \mathbf{b}$, where

Kf - the cryoscopic constant - supported the solvent
b - themolality of the solution;
i - the Van't Hoff factor - the amount of ions per individual molecule ofsolute.
$\mathbf{\Delta }\mathbf{T}\mathbf{f}\mathbf{r}\mathbf{e}\mathbf{e}\mathbf{z}\mathbf{i}\mathbf{n}\mathbf{g}$- The definition of temperature depression is

$\mathbf{T}\mathbf{F}\mathbf{(}\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{v}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{)}\mathbf{-}\mathbf{-}\mathbf{T}\mathbf{F}\mathbf{(}\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{)}$

As you’ll be able to see,$\mathbf{\Delta }\mathbf{T}\mathbf{f}\mathbf{r}\mathbf{e}\mathbf{e}\mathbf{z}\mathbf{i}\mathbf{n}\mathbf{g}$ two identicalsolutions that have different Van't Hoff factors are going to be bigger in favour of the one that dissociates into more ions.

In your case,

Salt contains a Van't Hoff factor of$\mathbf{i}\mathbf{=}2$

Glycerine includes a Van't Hoff factor of $\mathbf{i}\mathbf{=}\mathbf{1}\mathbf{.8}$