Question

Given,45 g of ethylene glycol (${{C}_{2}}{{H}_{6}}{{O}_{2}}$ ) is mixed with 600 g of water. Calculate a) freezing point depression and b) the freezing point of solution $[{{K}_{f}}$ for water = 1.86 K kg/mol, freezing point of water = 273 K].

Answer 1

Hint: There is a formula to calculate the depression in freezing point and it is as follows.
\[\Delta {{T}_{f}}={{K}_{f}}m\]
$\Delta {{T}_{f}}$ = depression in freezing point
${{K}_{f}}$ = freezing point depression constant
m = molality of the solution

Complete answer:
- In the question it is given that 45 g of ethylene glycol (${{C}_{2}}{{H}_{6}}{{O}_{2}}$) is mixed with 600 g of water.
- After adding ethylene glycol there is a depression in the freezing point of the solution and we have to calculate it.
- To calculate the depression in freezing point first we should know the molality of the solution.
- The formula to calculate the molality is as follows.
\[m=\dfrac{W}{M.W}\times \dfrac{1000}{Weight\text{ }of\text{ }the\text{ }solvent}\]
W = weight of the ethylene glycol = 45 g
M.W = molecular weight of the ethylene glycol = 62
 - Substitute all the known values in the above equation to get the molality of the solution and it is as follows.
\[\begin{align}
  & m=\dfrac{W}{M.W}\times \dfrac{1000}{Weight\text{ }of\text{ }the\text{ }solvent} \\
 & m=\dfrac{45}{62}\times \dfrac{1000}{600} \\
 & m=0.7258\times 1.66 \\
 & m=1.20m \\
\end{align}\]
a) Depression in freezing point
\[\Delta {{T}_{f}}={{K}_{f}}m\]
$\Delta {{T}_{f}}$ = depression in freezing point
${{K}_{f}}$ = freezing point depression constant = 1.86 K kg/mol
 m = molality of the solution = 1.20
- Substitute all the known values in the above formula to calculate the depression in freezing point.
\[\begin{align}
  & \Delta {{T}_{f}}={{K}_{f}}m \\
 & =1.86\times 1.20 \\
 & =2.25K \\
\end{align}\]
b) The freezing point of solution
 - There is a formula to calculate the freezing point of the solution and it is as follows.
\[\Delta T=T_{f}^{o}-{{T}_{f}}\]
$\Delta T$ = depression in freezing point = 2.25 K
$T_{f}^{o}$ = freezing point of water = 273 K
${{T}_{f}}$ = The freezing point of solution
- Substitute all the known values in the above formula to get the freezing point of solution and it is as follows.
\[\begin{align}
  & \Delta T=T_{f}^{o}-{{T}_{f}} \\
 & 2.25=273-{{T}_{f}} \\
 & {{T}_{f}}=270.75K \\
\end{align}\]

Note:
After adding some amount of ethylene glycol (solute) to the solvent (water) the freezing point of the water is going to decrease and it is called depression in the freezing point of the solution. The depression in freezing point is due to the addition of ethylene glycol.