Question

Ethylene glycol is used as an antifreeze agent. Calculate the amount of ethylene glycol used to be added to ${ 4kg }$ of water to prevent it from freezing at ${ -6 }^{ \circ }{ C }$
${ K }_{ f }$ for water = ${ 1.85Kmol }^{ -1 }{ Kg }$

Answer 1

Hint: Freezing point: It is defined as the temperature at which a liquid turns into a solid when cooled is called the freezing point of that particular matter.
${ \Delta T }_{ f }{ =1000\times K }_{ f }{ \times w }_{ 1 }{ /M/\times w }_{ 2 }$

Complete step-by-step answer:
 Freezing point depression is a colligative property seen in solutions that outcomes from the introduction of solute molecules to a solvent. The freezing points of solutions are all less than that of the pure solvent and are directly proportional to the molality of the solute.

It is given that;
W = ${ 4Kg = 4000g }$
${ K }_{ f }$ for water = ${ 1.85Kmol }^{ -1 }{ Kg }$
The freezing point of pure water = ${ 0 }^{ \circ }{ C }$
As we know,
${ \Delta T }_{ f }$ = freezing point of water - freezing point of ethylene glycol solution
${ \Delta T }_{ f }$ = ${ 0-\left( -6 \right) }^{ \circ }{ C }$ = ${ 6 }^{ \circ }{ C }$
The molecular mass of ethylene glycol = ${ 62gmol }^{ -1 }$
Using this formula, we get
${ \Delta T }_{ f }{ =1000\times K }_{ f }{ \times w }_{ 1 }{ /M/\times w }_{ 2 }$
${ w }_{ 1 }{ =\Delta T }_{ f }{ \times M\times }{ w }_{ 2 }{ /1000\times K }_{ f }$ …………….. (1)
where ${ w }_{ 1 }$ = mass of ethylene glycol in grams.
${ w }_{ 2 }$ = mass of solvent (water) in grams
M = molar mass of ethylene glycol
${ \Delta T }_{ f }$ = depression in freezing point
${ K }_{ f }$ = molar freezing point depression constant

Now, put the values in equation (1), we get
${ w }_{ 1 }{ = }{ 62\times 4000\times 6/1000\times 1.85 }$
         = ${ 804.32g }$.
Hence, the amount of ethylene glycol used = ${ 804.32g }$.

Additional Information:
Colligative properties are properties of solutions that rely on the proportion of the number of solute particles to the number of solvent molecules in a solution.
Colligative properties include:
1. Relative lowering of vapor pressure
2. Elevation of boiling point
3. Depression of freezing point
4. Osmotic pressure

Note: The possibility of making a mistake is that the weight of the solvent is given in kilograms, so you have to convert it to grams. Also, depression in the freezing point is equal to the freezing point of the solvent - freezing point of solute.