Question

In winter, the normal temperature in Kullu valley was found to be $- 11^\circ C$ . Is $28\%$ (by mass) aqueous solution of ethylene glycol suitable for car radiator?
${K_f}$ for water = $1 \cdot 86gmo{l^{ - 1}}$.

Answer 1

Depression in freezing is a phenomenon which means, after adding a solute to a solvent, the freezing point of the pure solvent in the solution will low down, and now the pure solvent will freeze at a lower temperature than earlier.

Complete step by step answer: The depression in freezing point can be calculated using the formula:
$\Delta T = m \times {K_f}$ ……….. $\left( 1 \right)$
where, $\Delta T$ = Change in the freezing point of the solvent,
$m$ = molality of the solution
${K_f}$ = Molal depression constant of water = $1 \cdot 86gmo{l^{ - 1}}$ (given)
Also, $\Delta T = {T_f}^\circ - {T_f}.........(2)$
where, ${T_f}^\circ$ = Initial freezing point of the pure solvent = $0^\circ C$ in case of water.
${T_f}$ = Freezing point of the pure solvent after adding the solute (to be calculated).
Molality, $m$, is defined as moles of solute per ${\text{kg}}$ of solvent.
So, $m = {\text{ }}\dfrac{{{\text{Moles of solute}}}}{{{\text{Mass of solvent in kg}}}}$ ………. $\left( 3 \right)$
Now, moles of solute can be calculated using the formula:
Moles of solute = $\dfrac{{{\text{Mass of solute}}}}{{{\text{Molar mass of solute}}}}$ ………… $\left( 4 \right)$
$28\%$ (by mass) aqueous solution of ethylene glycol is given in the question. That means, solute is ethylene glycol and solvent is water.
Let, the mass of $100{\text{ g}}$solution contains $28{\text{ g}}$ of solute.
Then mass of solution = $100{\text{ g}}$,
Mass of solute = $28{\text{ g}}$,
So, the mass of the solvent will be = $\left( {100 - 28} \right) = 72{\text{ g}}$.
Formula of ethylene glycol is $OH - C{H_2} - C{H_2} - OH$, so, its molar mass will be: $= \left( {2 \times {\text{mass of Carbon}} \times {\text{ }}2 \times {\text{Mass of Oxygen }} \times 6 \times {\text{Mass of Hydrogen}}} \right)$
$= 2 \times 12 + 2 \times 16 + 6 \times 1 = 62{\text{ g}}$
Now, from the equation $\left( 4 \right)$, moles of solute = $\dfrac{{28}}{{62}}mol$
From the equation $\left( 3 \right)$, molality of the solution, $m = \dfrac{{28}}{{62}} \times \dfrac{{1000}}{{72}} = 6 \cdot 27molk{g^{ - 1}}$,
Now, putting the value of $m$ in the equation $\left( 1 \right)$, $\Delta T = m \times {K_f}$ = $6 \cdot 27 \times 1 \cdot 86 = 11 \cdot 66^\circ C$
Finally, putting the value of $\Delta T$ in equation $\left( 2 \right)$, $\Delta T = {T_f}^\circ - {T_f}$ = $11.66^\circ C = 0^\circ C - {T_f}$,
${T_f} = 0 - 11.66 = - 11.66^\circ C$, which is the new freezing point of the solvent in the solution of ethylene glycol. It means that, now, the solvent (water in this case), will not freeze at $0^\circ C$ but will freeze at $- 11.66^\circ C$.
So, yes, a $28\%$ (by mass) aqueous solution of ethylene glycol is suitable for car radiator at the temperature of $- 11^\circ C$ in Kullu valley.

Note: In calculating molality, be careful while taking the masses, do not take the mass of the solution in place of the solvent in the denominator. Also, take the mass of the solvent in ${\text{kg}}$ only. The only difference between molality and the molarity is that in molality denominators have volume of solution but molality has mass of solvent in kilograms.