Question

Which has highest freezing point
A. ${\text{1}}$ m ${{\text{K}}_{\text{4}}}{\text{[Fe(CN}}{{\text{)}}_{\text{6}}}{\text{]}}$ solution
B. ${\text{1}}$ m ${{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}$ solution
C. ${\text{1}}$ m ${\text{KCl}}$ solution
D. ${\text{1}}$ m rock salt solution

Answer 1

Hint: The depression in freezing point is the product of the freezing point depression constant, Van’t Hoff factor and molality. The Van't Hoff factor is determined as the number of ions produced by the ionic compound in the solution. The depression in freezing point is directly proportional to the van’t Hoff factor. As the value of van’t Hoff factor increases depression in freezing point increases means the freezing point of the solution decreases.

Complete step by step solution:
The formula of freezing point depression is as follows:
${\Delta }{{\text{T}}_{\text{f}}}\,{\text{ = i}}\,{{\text{K}}_{\text{f}}}{\text{.}}\,{\text{m}}$
Where,
${\Delta }{{\text{T}}_{\text{f}}}$is the depression in freezing point.
${{\text{K}}_{\text{f}}}$is the freezing point depression constant.
${\text{i}}\,$is the van’t Hoff factor.
${\text{m}}$is the molal concentration
If the solvent is water then the freezing point depression constant of water is the same for all given solutions and molality is same for all the given solutions. So, we will compare the depression in freezing point on the basis of van’t Hoff factor.
The solution having the lowest value of the van't Hoff factor will have the highest freezing point.
${\text{1}}$ m ${{\text{K}}_{\text{4}}}{\text{[Fe(CN}}{{\text{)}}_{\text{6}}}{\text{]}}$ solution:
${{\text{K}}_{\text{4}}}{\text{[Fe(CN}}{{\text{)}}_{\text{6}}}{\text{]}}$is an ionic compound which dissociates in water as follows:
${{\text{K}}_{\text{4}}}{\text{[Fe(CN}}{{\text{)}}_{\text{6}}}{\text{]}}\, \to 4{{\text{K}}^ + }\, + \,{{\text{[Fe(CN}}{{\text{)}}_{\text{6}}}{\text{]}}^{2 - }}$
So, the value of van’t Hoff factor for ${{\text{K}}_{\text{4}}}{\text{[Fe(CN}}{{\text{)}}_{\text{6}}}{\text{]}}$ is $5$.
${\text{1}}$ m ${{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}$ solution:
${{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}$is an non-ionic compound so, it does not dissociates in water so, the value of van’t Hoff factor for ${{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}$ is $1$.
${\text{1}}$ m ${\text{KCl}}$ solution:
${\text{KCl}}$is an ionic compound which dissociates in water as follows:
${\text{KCl}}\, \to {{\text{K}}^ + }\, + \,{\text{C}}{{\text{l}}^ - }$
So, the value of the van't Hoff factor for ${\text{KCl}}$ is $2$.
${\text{1}}$ m rock salt solution:
${\text{NaCl}}$is an ionic compound which dissociates in water as follows:
${\text{NaCl}} \to \,{\text{N}}{{\text{a}}^{\text{ + }}}\, + \,{\text{C}}{{\text{l}}^ - }$
So, the value of van’t Hoff factor for ${\text{NaCl}}$ is$2$.
So, the van’t Hoff factor is lowest for ${{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}$ solution so, the ${\text{1}}$ m ${{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}$ solution has highest freezing point.
Therefore, option (C) ${\text{1}}$ m${{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}$ solution, is correct.

Note:When a non-volatile solute is added to the pure solvent, the vapour pressure of the solution decreases. So, the freezing point of the solution decreases which is known as the depression in freezing point. The freezing point of the solution will be less than the freezing point of the solvent. The freezing point of the pure solvent at a temperature is defined as the freezing point constant. The Van't Hoff factor represents the degree of dissociation, association or number of ions produced by a compound on dissolution.