Question

An antifreeze solution is prepared by dissolving $31g$ of ethylene glycol (${C_2}{H_6}{O_2}$), in $600g$ of water. Calculate the freezing point of the solution (${K_f}$ for water $ = 1.86Kmo{l^{ - 1}}k{g^{ - 1}}$).

Answer 1

Hint: First, we can find the molality of the given solution using the data given. We can then use this to find the depression in freezing point, which is equal to the product of molality and the cryoscopic constant. Subtracting the value of depression of freezing point from the actual freezing point of water will give us the freezing point of the solution.

Formulas used:
$\Delta {T_f} = m{K_f}$
Where $\Delta {T_f}$ is the depression in freezing point, $m$ is the molality of the solution and ${K_f}$ is the cryoscopic constant.
$m = \dfrac{{{n_B}}}{{{W_A}}} \times 1000$
Where ${n_B}$ is the number of moles of the solute and ${W_A}$ is the weight of the solvent in grams.
${n_B} = \dfrac{{{W_B}}}{{{M_B}}}$
Where ${W_B}$ is the given mass of the solute and ${M_B}$ is the molar mass of the solute.

Complete step by step answer:
When a solute is dissolved in water, the freezing point of water reduces, and the amount of reduction is given by the formula:
$\Delta {T_f} = m{K_f}$
Where $\Delta {T_f}$ is the depression in freezing point, $m$ is the molality of the solution and ${K_f}$ is the cryoscopic constant.
Thus, we need to first find the molality of the solution, given by the formula:
$m = \dfrac{{{n_B}}}{{{W_A}}} \times 1000$
Where ${n_B}$ is the number of moles of the solute and ${W_A}$ is the weight of the solvent in grams.
As we know, to find the number of moles:
${n_B} = \dfrac{{{W_B}}}{{{M_B}}}$
Where ${W_B}$ is the given mass of the solute and ${M_B}$ is the molar mass of the solute.
Here the given mass of solute is $31g$. The molar mass of carbon $ = 12g$, hydrogen $ = 1g$ and oxygen $ = 16g$. Thus, the molecular mass of ethylene glycol (${C_2}{H_6}{O_2}$) is:
$(12 \times 2) + (1 \times 6) + (16 \times 2) = 62g/mol$. Hence, the number of moles of (${C_2}{H_6}{O_2}$) is:
${n_B} = \dfrac{{31}}{{62}} = 0.5mol$
The mass of solvent given is $600g$. Substituting this and ${n_B} = 0.5$, we get the molality as:
$m = \dfrac{{0.5}}{{600}} \times 1000$
$ \Rightarrow {m_B} = \dfrac{{500}}{{600}} = 0.833m$
Substituting this value and ${K_f} = 1.86$ into the freezing point equation, we get the depression in freezing point as:
$\Delta {T_f} = 0.833 \times 1.86$
$ \Rightarrow \Delta {T_f} = 1.55K$
As we know, the freezing point of water is $0^\circ C = 273.15K$
$F.{P_{solution}} = F.{P_{solvent}} - \Delta {T_f}$
Where $F.{P_{solution}}$ is the freezing point of the solution and $F.{P_{solvent}}$ is the freezing point of the pure solvent. Substituting the values, we get:
$F.{P_{solution}} = 273.15 - 1.55 = 271.6K$

Hence, the freezing point of the solution is $271.6K$

Note: When a solute is added to a solvent, not only does its freezing point get reduced, but its boiling point is also elevated. This is caused due to the interaction between the solute and solvent particles. Correspondingly, the constant present in the elevation of boiling point equation is known as the ebullioscopic constant (${K_b}$).