Question

A car is participating in Himalayan car Rally. The temperature of the valley is $-6{}^{0}C$. What amount of ethylene glycol should be added to water in a car radiator to prevent its freezing? The radiator contains 6kg water. Due to a sudden thunderstorm, the temperature of the valley was further lowered to $-8{}^{0}C$. What amount of ice will separate out? ${{K}_{f}}$ for water is 1.86 K-kg/mol.

Answer 1

freezing point of a liquid is the temperature at which vapor pressure of liquid phase of substance becomes equal to the vapor pressure of solid phase of the substance. Addition of any non-volatile solute into a solvent decreases its freezing point because its addition decreases the vapor pressure, therefore now liquid solvent has to be taken to lower temperature to bring its vapor pressure equal to the vapor pressure of solid solvent.

Complete step by step solution:
To calculate the depression in freezing point we will use the formula shown below:
$\Delta {{T}_{f}}={{K}_{f}}\times \dfrac{{{w}_{A}}\times 1000}{{{M}_{A}}\times {{w}_{B}}}$
$\Delta {{T}_{f}}$= depression in freezing point
${{K}_{f}}$= Molal depression constant
${{w}_{A}}$=Weight of the non-volatile solute dissolved in gram
${{M}_{A}}$=Gram molecular weight of non-volatile solute
${{w}_{B}}$=Weight of the solvent in Kg
Since freezing point of water is ${{0}^{0}}C$ therefore depression in freezing point will be ${{6}^{0}}C$ as freezing point after depression is $-{{6}^{0}}C$
$6=1.86\times \dfrac{{{w}_{A}}\times 1000}{62\times 6}$
$6=1.86\times \dfrac{{{w}_{A}}\times 1000}{372}$
$6\times 372=1860\times {{w}_{A}}$
$2232=1860\times {{w}_{A}}$
$\dfrac{2232}{1860}={{w}_{A}}$
${{w}_{A}}=\dfrac{2232}{1860}$
$\therefore {{w}_{A}}=1.2g$
Therefore any quantity of ethylene glycol more than 1.2 g should be added in the car radiator to stop the water from freezing.
If the temperature of the valley is $-8{}^{0}C$ then let’s calculate the amount of water that will dissolve 1.2 g ethylene glycol
Since freezing point of water is ${{0}^{0}}C$ therefore depression in freezing point will be ${{8}^{0}}C$ as freezing point after depression is $-{{8}^{0}}C$
$\Delta {{T}_{f}}={{K}_{f}}\times \dfrac{{{w}_{A}}\times 1000}{{{M}_{A}}\times {{w}_{B}}}$
$8=1.86\times \dfrac{1.2\times 1000}{62\times {{w}_{B}}}$
$8=1.86\times \dfrac{1200}{62\times {{w}_{B}}}$
$8\times 62\times {{w}_{B}}=2232$
$496\times {{w}_{B}}=2232$
$\therefore {{w}_{B}}=\dfrac{2232}{496}$
${{w}_{A}}=4.5kg$
Therefore 4.5 kg of water will be good enough to dissolve 1.2 g ethylene glycol at the temperature of $-8{}^{0}C$ and rest of the water will separate out as ice.
Amount of water separated out as ice= 6-4.5
Amount of water separated out as ice= 1.5Kg

Additional information:
Vapor pressure is defined as the pressure exerted by a vapor on the surface of the liquid that is in equilibrium with it.

Note: A solution is always a homogeneous mixture, therefore when we dissolve a non-volatile solute into a solvent; some of the solute particles displace the solvent particle from the surface. As we know that there is always a dynamic equilibrium at the surface of a liquid where condensation and evaporation keeps happening continuously. So now there are less particles of solvent available on the surface to take part in equilibrium, therefore vapor pressure will decrease.