Question

Ethylene glycol is used as antifreeze to reduce freezing point of water to $-{{2.4}^{\circ }}C$. What mass of antifreeze is required for $2L$water? $({{K}_{f}}water=1.86\dfrac{Kkg}{mole})$
A. $16kg$
B. $160g$
C. $1.60kg$
D. $16g$

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. We know that depression in the freezing point of a matter is given by the formula $\Delta {{T}_{f}}=i\times {{K}_{f}}\times m$.

Complete step-by-step answer:
The property of decrease in freezing point of a solvent (here water) when sone non-volatile solute (here ethylene glycol) is dissolved is called depression of freezing point. The depression in freezing point is represented by $\Delta {{T}_{f}}$.
In the question it is given to us that
$T_{f}^{'}=-{{2.4}^{\circ }}C$ (new freezing point obtained after addition of solute)
${{T}_{f}}={{0}^{\circ }}C$(actual freezing point of water)
${{K}_{f}}=1.86\dfrac{Kkg}{mole}$
Volume of water = $2L$
We can calculate $\Delta {{T}_{f}}={{T}_{f}}-T_{f}^{'}=0-\left( -2.4 \right)={{2.4}^{\circ }}C$
Now, we know that the depression in freezing point is given by the formula
$
 \Delta {{T}_{f}}=i\times {{K}_{f}}\times m \\
  i=1 \\
$
Therefore, $\Delta {{T}_{f}}={{K}_{f}}\times m$… (equation 1)
Substituting, the given values in the equation 1, we get
$
 \Rightarrow 2.4=1.86\times m \\
 \Rightarrow m=\dfrac{2.4}{1.86} \\
 \Rightarrow m=1.3 \\
$
Therefore, molality of the solution is m= 1.3

Now, we know that molality is defined as the moles of solute dissolved per kg of solvent
Or, we can say that molality (m) = $\dfrac{moles}{2}$
Because, mass of 2L water is 2Kg
Therefore, moles= molality $\times 2$
Moles = $1.3\times 2=2.6$moles
Now, we know that mole of a compound is given by the formula,
Mole = $\dfrac{mass}{molar\,\,mass}$
Therefore, mass = mole $\times $molar mass
Mass of ethylene glycol can be calculated as follows
$
 2.6\times 62.07 \\
 \Rightarrow 160g \\
$
Therefore, the mass of ethylene glycol needed is $160 g$

Hence the correct answer is option ‘B’.

Note: It must be clear to the students that the solution will be non-volatile. Calculations must be done carefully and if the question asks about the freezing point of solution then the reference is taken to be water. Water is taken as universal solvent and its freezing point is zero degree Celsius.