Question

An aqueous solution freezes at -2.55 $^{0}C$ what is the boiling point ($K_{b}^{{{H}_{2}}O}=0.52K{{m}^{-1}},K_{f}^{{{H}_{2}}O}=1.86K{{m}^{-1}}$ )?
(a) 107.0$^{0}C$
(b) 100.6$^{0}C$
(c) 100.1$^{0}C$
(d) 100.7$^{0}C$

Answer 1

Hint: The freezing point is the temperature at which the solvent in the liquid state and the solvent in the solid state are present at equilibrium so that their vapour pressure becomes equal.

Complete step by step solution:
Given in the question: $K_{b}^{{{H}_{2}}O}=0.52K{{m}^{-1}},K_{f}^{{{H}_{2}}O}=1.86K{{m}^{-1}}$
To solve this question first we have to find the value of m using $\Delta {{T}_{f}}$ and after the calculation of m we will calculate the value of $\Delta {{T}_{b}}$and then it is help to calculate the value of boiling point.
First step :
Calculation of $\Delta {{T}_{f}}$ = ${{k}_{f}}m$
And m = $\dfrac{\Delta {{T}_{f}}}{{{k}_{f}}}$ =$\dfrac{2.55}{1.86}$
Second step:
$\Delta {{T}_{b}}$= ${{k}_{b}}m$
$\implies$ \[0.52(\dfrac{02.55}{1.86})=0.71\]
The value of ${{T}_{BP}}$ = $100+0.71$ = 100.7 $^{0}C$

Hence the correct answer is option (D) i.e. an aqueous solution freezes at -2.55 $^{0}C$ the boiling point is 100.7$^{0}C$.

Additional information:
Colligative properties are the properties of a solution which solely depends on the number of particles of the solute and not the type of nature of the solute. Freezing point of water is one of the colligative properties.

Note: Do not get confused between the terms depression in freezing point and freezing point. Freezing point is the temperature at which the solvent in the liquid state and the solvent in the solid state are present in equilibrium and the depression in freezing point is the difference in the freezing point of the solution from its pure solvents. This is true that if any solute is added to a solvent the freezing point of the solution will be lesser than the freezing point of the pure solvent.