Question

A solution of sucrose (molar mass \[ = 342{\text{ }}g{\text{ }}mo{l^{ - 1}}\]) has been prepared by dissolving \[68.5{\text{ }}g\]of sucrose in \[1000{\text{ }}g\]of water. The freezing point of the solution obtained will be:
[\[{K_f}\] for water \[ = 1.86{\text{ }}K{\text{ }}kg{\text{ }}mo{l^{ - 1}}\]]
A. \[ - 0.372^\circ C\]
B. \[0.520^\circ C\]
C. \[ + 0.372^\circ C\]
D. \[0.570^\circ C\]

Answer 1

Hint: Here, in this question, a solution is prepared by dissolving \[68.5{\text{ }}g\]of sucrose in \[1000{\text{ }}g\]of water. We have to identify the freezing point of the solution of sucrose.

Complete step by step solution:
The phenomenon of lowering the freezing point of any solvent after the addition of solutes is known as freezing point depression. It should be evident from this description that freezing point depression is a colligative feature of solutions. The freezing point formula is usually proportional to the molality of the solute that is added in the majority of circumstances. With this knowledge, it is also possible to deduce that the freezing point formula is:
\[\Delta {T_f} = i \times {K_f} \times m\]
Here, \[\Delta {T_f}\] is freezing point depression, \[i\] is van’t Hoff factor, \[{K_f}\] is freezing point depression constant, and \[m\] is the molarity.
We can convert the above equation as follows:
\[\Delta {T_f} = \dfrac{{{W_1} \times {K_f} \times 1000}}{{{M_1} \times {W_2}}}\]
Here, \[{W_1}\] is weight of solute, \[{W_2}\] is weight of solvent, \[{M_1}\] is molar mass of solute, and \[{K_f}\] is freezing point depression constant.
Let’s substitute the value in the formula,
\[
  \Delta {T_f} = \dfrac{{{W_1} \times {K_f} \times 1000}}{{{M_1} \times {W_2}}} \\
   \Rightarrow \Delta {T_f} = \dfrac{{1.86 \times 68.5 \times 1000}}{{342 \times 1000}} \\
   \Rightarrow \Delta {T_f} = 0.372^\circ C \\
 \]
Now, calculate freezing point of solute as follows:
\[\Delta {T_f} = T^\circ - {T_f}\]
Here, \[\Delta {T_f}\] is depression in freezing point, \[T^\circ \] is freezing point of pure water, and \[{T_f}\] is freezing point of solution.
Let us calculate freezing point of solute,
\[
  \Delta {T_f} = T^\circ - {T_f} \\
   \Rightarrow {T_f} = T^\circ - \Delta {T_f} \\
 \]
Substitute the value and solving it,
\[
  {T_f} = T^\circ - \Delta {T_f} \\
   \Rightarrow {T_f} = 0 - 0.372^\circ C \\
   \Rightarrow {T_f} = - 0.372^\circ C \\
 \]
As a result, the freezing point of the solution is \[ - 0.372^\circ C\].
Therefore, the correct answer is option A \[ - 0.372^\circ C\].
Additional Information: Seawater has a freezing point of below zero degrees Celsius. At temperatures below the freezing point of pure water, seawater remains liquid. The salts dissolved in the ocean are to blame for this. A common example of this behaviour can be seen in an ethanol-in-water solution. The solution freezes at a lower temperature than pure water but at a greater temperature than pure ethanol.
Note: At the freezing point of a solvent, there is a condition of equilibrium between the solid and liquid states of the solvent. This means that the solid and liquid phases have the same vapour pressure. The solid and the solution eventually attain an equilibrium at lower temperatures as a result of the entire procedure.