Question

On February 14,2003 the temperature In Manali was found to be${\text{ - 10}}^\circ {\text{C}}$ . A car was used, whose radiator was filled with 5L of water. What quantity of anti freezing agent ethylene glycol were added to water of radiator in order to use the car for travelling (given ${{\text{K}}_{\text{f}}}$ of water =$1.86{\text{K kg mol}}{{\text{l}}^{{\text{ - 1}}}}$, $d = {\text{1 g m}}{{\text{l}}^{{\text{ - 1}}}}$)
A) 3200 g
B) 1670 g
C) 3000 g
D) 2100 g

Answer 1

Using the depression of the freezing point equation calculate the molality of the solution. Using the density and volume of water calculate the mass of water. Using the mass of water and molality of solution calculate moles of solute ethylene glycol. Finally, convert the moles of ethylene glycol to the mass of it using its molar mass.

Complete solution:
The freezing point is the temperature at which the liquid and solid state of substance have the same vapour pressure.
The freezing point of pure water is ${\text{0}}^\circ {\text{C}}$. The temperature in Manali was found to be${\text{ - 10}}^\circ {\text{C}}$ which caused freezing of water in the radiator. The addition of anti-freezing agent ethylene glycol causes depression in the freezing point of water.
Now, using the freezing point of pure solvent water and freezing point of the solution we can calculate the depression of the freezing point of solution($\Delta {T_f}$).
$\Delta {T_f} = {\text{freezing point of pure solvent - freezing point of solution}}$
$\Delta {T_f} = {\text{0}}^\circ {\text{C}} - ( - 10^\circ {\text{C}}) = 10^\circ {\text{C}}$
Now, using this calculated value of depression of freezing point of the solution and molal freezing point depression constant ${{\text{K}}_{\text{f}}}$ of water ($1.86{\text{K kg mol}}{{\text{l}}^{{\text{ - 1}}}}$) we can calculate the molal concentration (${C_m}$) of the solution as follows:
$\Delta {T_f} = {{\text{K}}_{\text{f}}} \times {C_m}$
${C_m} = \dfrac{{\Delta {T_f}}}{{{{\text{K}}_{\text{f}}}}} = \dfrac{{{\text{10}}^\circ {\text{C}}}}{{1.86{\text{K kg mol}}{{\text{l}}^{{\text{ - 1}}}}}} = 5.38{\text{mol k}}{{\text{g}}^{{\text{ - 1}}}}$
Now, using the given density of water and volume of water we can calculate the mass of water as follows:
$d = \dfrac{m}{v}$
Volume of water = 5 L = 5000ml
Now, substitute ${\text{1 g m}}{{\text{l}}^{{\text{ - 1}}}}$ for the density of water and 5000ml for the volume of water and calculate the mass of water.
$m = d \times v$
$m = 1{\text{ g m}}{{\text{l}}^{{\text{ - 1 }}}} \times 5000{\text{ ml = 5000 g = 5kg}}$
Now, we have the molality of solution and mass of solvent so we can calculate the moles of solute ethylene glycol as follows:
${\text{molality = }}\dfrac{{{\text{moles of solute }}}}{{{\text{Kg of solvent}}}}$
${\text{moles of solute}} = {\text{molality }} \times {\text{Kg of solvent}}$
${\text{moles of ethylene glycol}} = 5.38{\text{mol k}}{{\text{g}}^{{\text{ - 1}}}} \times 5{\text{Kg = 26}}{\text{.9 mol}}$
Now, using the molar mass of ethylene glycol and moles of it we can calculate the mass of ethylene glycol.
The molar mass of ethylene glycol = $62.07{\text{ g mo}}{{\text{l}}^{{\text{ - 1}}}}$
${\text{Mass = mol }} \times {\text{ molar mass}}$
${\text{Mass = 26}}{\text{.9 mol }} \times {\text{ }}62.07{\text{ g mo}}{{\text{l}}^{{\text{ - 1}}}} = 1670{\text{g}}$
Thus, 1670 g anti freezing agent ethylene glycol was added to the water of the radiator in order to use the car for travelling.

Hence, the correct answer is an option (B) 1670 g.

Note: Depression of freezing point is the colligative property of the solution. A property that does not depend on the nature of solute but depends upon the concentration of solute present in the solution is known as the colligative property of the solution.