Question

An aqueous solution freezes at $- {2.55^\circ }{\text{C}}$. What is its boiling point?
$\left[ {{\mathbf{K}}_{\text{b}}^{{{\text{H}}_2}{\text{O}}} = 0.52\;{\text{K}}\;{{\text{m}}^{ - 1}};{\mathbf{K}}_1^{{{\text{H}}_2}{\text{O}}} = 1.86\;{\text{K}}\;{{\text{m}}^{ - 1}}} \right]$,
A ${107.0^\circ }{\text{C}}$
B ${100.6^\circ }{\text{C}}$
C ${100.1^\circ }{\text{C}}$
D ${100.7^\circ }{\text{C}}$

Answer 1

A substance's boiling point is the temperature at which the vapour pressure of a liquid equals the pressure around the liquid and the liquid transforms into a vapour. The boiling point of a liquid is affected by the surrounding atmospheric pressure. The boiling point of a liquid in a partial vacuum is lower than the boiling point of the same liquid at atmospheric pressure. The boiling point of a liquid at high pressure is greater than the boiling point of the same liquid at atmospheric pressure.

Complete answer:
Delta(T) = Km is used to determine the freezing-point depression, where K is the solvent's freezing-point depression constant and m is the molality of the solution. In this example, molality refers to the number of moles of solute particles per kg of solvent.
A substance's melting point is the temperature at which it transitions from solid to liquid condition. The solid and liquid phases are in balance at the melting point. A substance's melting point is determined by pressure and is generally given at a standard pressure such as 1 atmosphere or 100 kPa. The freezing point, also known as the crystallisation point, is the temperature at which a liquid changes back into a solid. The freezing point of a material might readily appear to be lower than its true value due to the tendency of substances to supercool.
Now we know that
$\Delta {{\text{T}}_{\text{f}}} = {{\text{K}}_{\text{f}}}{\text{m}}$
Upon substitution we get
${\text{m}} = \dfrac{{\Delta {{\text{T}}_{\text{f}}}}}{{{{\text{K}}_{\text{f}}}}} = \left( {\dfrac{{2.55}}{{1.86}}} \right)$
also
$\Delta {{\text{T}}_{\text{b}}} = {{\text{K}}_{\text{b}}}{\text{m}}$
Upon substitution we get
$= 0.52\left( {\dfrac{{02.55}}{{1.86}}} \right)$
=0.71
So $\Delta {{\text{T}}_{\text{b}}} = {0.71^\circ }{\text{C}}$
${{\text{T}}_{{\text{B}}.{\text{P}}}} = 100 + {0.71^\circ }{\text{C}}$
$\Rightarrow {T_B} = {100.7^\circ }{\text{C}}$
Hence, the correct option is D.

Note:
For a given solvent, Kf is a constant. When 1.00 mole of a nonvolatile nonionizing (non dissociating) solute dissolves in one kilogramme of solvent, Kf (molal freezing point depression constant) reflects how many degrees the freezing point of the solvent changes. The constants Kb and Kf are used to determine the variations in boiling and freezing points seen in nonvolatile nonelectrolyte solute solutions. (a) are affected by the solute's concentration. (b) are determined by the solute's characteristics. (c) are affected by the solvent's characteristics.