Question

How many grams of methyl alcohol should be added to \[10{{L}}\] tank of water to prevent its $268{{K}}$?
${{{K}}_{{f}}}$ of water is $1.84K.g.mol^{- 1}$.
A. $899.04{{g}}$
B. $886.02{{g}}$
C. $868.06{{g}}$
D. $880.07{{g}}$

Answer 1

Hint: The freezing point is the temperature at which a substance in liquid state changes to solid-state. Theoretically, when the vapor pressure of a liquid equals the vapor pressure of a solid. Moreover, the freezing point depends on the van’t Hoff factor and the molality.

Complete step by step answer:
The vapor pressure of a solution is decreased when a non-volatile solute is added to a volatile solvent. This decreases vapor pressure has many properties. Freezing point depression change is denoted by $\Delta {{{T}}_{{f}}}$.
\[\Delta {{{T}}_{{f}}} = {{{K}}_{{f}}}{{m}}\], where \[{{{K}}_{{f}}}\]is the freezing point depression constant.
\[{{m}}\]is the molality of the solution.
It is given that the freezing point depression constant of water, ${{{K}}_{{f}}} = 1.84{{K}}.{{kg}}.{{mo}}{{{l}}^{ - 1}}$
The volume of water, ${{{V}}_{{w}}} = 10{{L}}$
Initially, we have to calculate the mass of solvent, i.e. water. We know that the density of water, ${\rho _{{w}}} = 1{{kg}}{{{L}}^{ - 1}}$. Density is defined as mass by volume. Therefore mass of water can be calculated by multiplying density of water with volume of water.
i.e. Mass of water, ${{{m}}_{{w}}} = {\rho _w} \times {{{V}}_{{w}}}$
${{{m}}_{{w}}} = 1{{kg}}{{{L}}^{ - 1}} \times 10{{L = 10kg}}$
Freezing point depression change, $\Delta {{{T}}_{{f}}}$ is the difference of freezing point of pure solvent, ${{{T}}_{{f}}}^ \circ $ and freezing point of solution, ${{{T}}_{{f}}}$.
Freezing point of water, ${{{T}}_{{f}}}^ \circ $ is ${0^ \circ }{{C}}$, i.e. $273{{K}}$ and freezing point of solution, ${{{T}}_{{f}}}$ is given as $268{{K}}$.
Thus $\Delta {{{T}}_{{f}}} = {{{T}}_{{f}}}^ \circ - {{{T}}_{{f}}}$
$\Delta {{{T}}_{{f}}} = 273{{K}} - 268{{K}} = 5{{K}}$
Molality can be calculated by dividing the number of moles of solute in ${{kg}}$ of solvent. The number of moles is the mass of solute divided by the molar mass of solute.
Combining the above definitions, we get the formula of molality.
Molality, ${{m = }}\dfrac{{{{{m}}_{{s}}}}}{{{{{M}}_{{s}}} \times {{{m}}_{{w}}}}} \times 1000$, where ${{{M}}_{{s}}}$ is the molar mass of solute.
Substituting the values, we get

${{m = }}\dfrac{{{{{m}}_{{s}}}}}{{32 \times 10}} $
${{m = }}\dfrac{{{{{m}}_{{s}}}}}{{320}} $
Substituting all the values in the equation to find the lowering in freezing point, we get
\[5 = 1.84 \times \dfrac{{{{{m}}_{{s}}}}}{{320}}\]
\[5 = 5.75 \times {10^{ - 3}}{{{m}}_{{s}}}\]
Thus, ${{{m}}_{{s}}} = 869.56{{g}}$
Thus the mass of methyl alcohol will be $869.56{{g}}$
The number closest to the answer in the given options is $868.06{{g}}$

So, the correct answer is Option C .

Note:
When solutes are added to a solvent, forming a solution, solute molecules disrupt the formation of solvent’s crystals. This disruption in the freezing process results in a depression of freezing point for the solution compared to the solvent.