Question

Freezing point of an aqueous solution is \[\text{-0}\text{.186}{}^\circ \text{C}\]. If the values of ${{\text{K}}_{\text{b}}}\text{ and }{{\text{K}}_{\text{f}}}$ of water are respectively $\text{0}\text{.52 K kg mo}{{\text{l}}^{-1}}\text{and 1}\text{.86 K kg mo}{{\text{l}}^{-1}}$ then the elevation of boiling point of the solution in K is:
A. 0.52
B. 1.04
C. 1.34
D. 0.134

Answer 1

Hint: The freezing point is given whose expression or formula is $\vartriangle {{\text{T}}_{\text{f}}}\text{ = }{{\text{K}}_{f}}\text{ }\times \text{ m}$. Here, $\vartriangle {{\text{T}}_{\text{f}}}$ is a depression in the freezing point, m is the molality and ${{\text{K}}_{f}}$ is the freezing point depression constant. Then after finding the value of molality we can find the value of elevation of boiling point by using the formula $\vartriangle {{\text{T}}_{b}}\text{ = }{{\text{K}}_{b}}\text{ }\times \text{ m}$.

Complete Answer:
- In the given question, we have to find the value of elevation in the boiling point in the solution.
- It is given that the value of a freezing point i.e. $\vartriangle {{\text{T}}_{\text{f}}}$ is \[\text{-0}\text{.186}{}^\circ \text{C}\] and ${{\text{K}}_{f}}$ is equal to the $\text{1}\text{.86 K kg mo}{{\text{l}}^{-1}}$, so by applying the formula of depression in freezing point we will get:
  $\vartriangle {{\text{T}}_{\text{f}}}\text{ = }{{\text{K}}_{f}}\text{ }\times \text{ m}$
$\text{0}\text{.186 = 1}\text{.86 }\times \text{ m}$
  m = 0.1 mol/kg
- So, now we know the value of molality that is equal to 0.1 mol/kg.
- Now, it is given that the value ${{\text{K}}_{b}}$ or boiling point elevation constant is $\text{0}\text{.52 K kg mo}{{\text{l}}^{-1}}$.
- So, by applying the formula of elevation in the boiling point we will get the value of elevation in the boiling point in the solution i.e.
$\vartriangle {{\text{T}}_{b}}\text{ = }{{\text{K}}_{b}}\text{ }\times \text{ m}$
- By putting the value of elevation of the boiling point and boiling point elevation constant, we will get:
$\vartriangle {{\text{T}}_{\text{f}}}\text{ = 0}\text{.52 }\times \text{ 0}\text{.1 = 0}\text{.052}{}^\circ \text{C}$
- So, the elevation in the boiling point of the solution is $0.052{}^\circ \text{C}$.

Therefore, option D is the correct answer.

Note:
Molality is defined as the ratio of the number of the moles of solute to the volume of the solvent in kilograms. Depression in the freezing point and elevation in the boiling point is both colligative property (those who depend on the concentration of the solute).