Question

What mass of ethylene glycol (molar mass = 62.0 $g.mo{l^{ - 1}}$ ) must be added to 5.50 kg of water to lower the freezing point of water from ${0^ \circ }C$ to $- {10.0^ \circ }C$ ?
(${K_f}$ for water = 1.86 $K.kg.mo{l^{ - 1}}$ )

Answer 1

Hint : First calculate the depression in freezing point by subtracting the freezing point after adding the solute from the freezing point of water, then find out the molality which is a concentration term which will lead you to the mass of solute to be added.

Formulas used:
-Depression in freezing point: $\Delta {T_f} = {K_f}m$ (1)
Where, $\Delta {T_f} = {T_f}^ \circ - {T_f}$ (2)
${T_f}^ \circ$ = freezing point of pure solvent
${T_f}$ = freezing point when non-volatile solute is dissolved in it
m = molality
${K_f}$ = Freezing point depression constant or Molal depression constant or Cryoscopic constant
-Molality (m) = moles of solute / kg of solvent
$m = \dfrac{{{W_B}}}{{{M_B}}} \times \dfrac{{1000}}{{{W_A}}}$ (3)
${W_B}$ = given weight of solute (in g)
${M_B}$= molecular weight of solute (in g)
${W_A}$ = given weight of solvent (in g)

Complete step by step solution :
-The values given in the question are:
Molecular weight of solute (ethylene glycol) = 62.0 $g.mo{l^{ - 1}}$
Given weight of solvent (water) = 5.5 kg
${K_f}$ = 1.86 $K.kg.mo{l^{ - 1}}$
${T_f}^ \circ$ = ${0^ \circ }C$
${T_f}$= $- {10.0^ \circ }C$

- From equation (2) we can calculate the depression in freezing point:
$\Delta {T_f} = {T_f}^ \circ - {T_f}$
$\Delta {T_f}$ = 0 – (-10) = 0 + 10
= ${10^ \circ }C$
-Since we have the value of $\Delta {T_f}$ and ${K_f}$, we can calculate the value of molality and from that we can calculate the mass of ethylene glycol (solute) to be added to cause the depression in freezing point.
-From equation (1) we will calculate the value of molality: $\Delta {T_f} = {K_f}m$
10 = 1.86 × m
m = 10 / 1.86 = 5.37
The molality is 5.37 molal.
-From the value of molality we will now find out the mass of ethylene glycol (solute) to be added.

Using equation (3): $m = \dfrac{{{W_B}}}{{{M_B}}} \times \dfrac{{1000}}{{{W_A}}}$
${W_B}$= ? (What we need to find out)
${M_B}$= 62 $g.mo{l^{ - 1}}$
${W_A}$= 5.5 kg = 5500 g
$5.37 = \dfrac{{{W_B}}}{{62}} \times \dfrac{{1000}}{{5500}}$
${W_B} = \dfrac{{5.37 \times 62 \times 5500}}{{1000}}$
= 1831.77 g
So, 1831.77 g of ethylene glycol must be added to 5.5 kg of water to lower the freezing point of water from ${0^ \circ }C$ to $- {10.0^ \circ }C$.

Note :
The depression in freezing point ($\Delta {T_f}$) is directly proportional to the molality of the solution only if the solution is an ideal solution or dilute solution. This proportionality relation will not be valid if the solution is a non-ideal solution.
$\Delta {T_f}\alpha m$