Question

At ${100^ \circ }C$ the vapour pressure of a solution of $6.5g$ of a solute in $100g$ water is $732mm$. If ${k_b} = 0.52$, the boiling point of this solution will be:
A.${100^ \circ }C$
B.${102^ \circ }C$
C.${103^ \circ }C$
D.${101^ \circ }C$

Answer 1

Hint: This question gives the knowledge about elevation in boiling point. Elevation in boiling point is defined as the increase in the boiling point of the solvents when solutes are added into the solution. It is a colligative property.

Formula used:
The formula used to determine the relative lowering of vapour pressure is as follows:
$\dfrac{{P_A^ \circ - {P_A}}}{{{P_A}}} = \dfrac{{{n_{solute}}}}{{{n_{solvent}}}}$
Where $P_A^ \circ $ is the vapor pressure of pure solvent, ${P_A}$ is the vapor pressure of solution, ${n_{solute}}$ is the moles of solute and ${n_{solvent}}$ is the moles of solvent.
The formula used to determine the molality of the solution is as follows:
$m = \dfrac{n}{V} \times 1000$
Where $m$ is the molality of the solution, $n$ is the number of moles of solute and $V$ is the volume of the solvent in gram.
 The formula used to determine the elevation in boiling point is as follows:
$ \Rightarrow \Delta {T_b} = {k_b}.m$
Where $\Delta {T_b}$ is the elevation in boiling point, $m$ is the molality, ${k_b}$ is the ebullioscopic constant.

Complete step by step answer:
 The colligative properties generally depend upon the number of particles.
First we will determine the number of moles of solute using relative lowering of vapour pressure formula as follows:
$ \Rightarrow \dfrac{{P_A^ \circ - {P_A}}}{{{P_A}}} = \dfrac{{{n_{solute}}}}{{{n_{solvent}}}}$
Substitute $P_A^ \circ $ as $760$, ${P_A}$ as $732$, and ${n_{solvent}}$ as $55.5$in the above formula.
$ \Rightarrow \dfrac{{760 - 732}}{{732}} = \dfrac{{{n_{solute}}}}{{55.5}}$
On simplifying, we get
$ \Rightarrow {n_{solute}} = \dfrac{{28}}{{732}} \times 55.5$
On further simplifying, we get
$ \Rightarrow {n_{solute}} = 0.203$
So, the moles of solute are $0.203$.
Using moles of solute now we will determine the molality of the solvent.
$ \Rightarrow m = \dfrac{n}{V} \times 1000$
Substitute moles of solute as $0.203$ and volume as $100$ in the above formula.
$ \Rightarrow m = \dfrac{{0.203}}{{100}} \times 1000$
On simplifying, we get
$ \Rightarrow m = 0.203 \times 10$
On further simplifying we get
$ \Rightarrow m = 2.03$
So, the molality of the solution is $2.03$.
Elevation in boiling point is defined as the increase in the boiling point of the solvents when solutes are added into the solution. It is a colligative property.
 Using the formula of elevation in boiling point we will determine the boiling point as follows:
$ \Rightarrow \Delta {T_b} = {k_b}.m$
Substitute molality as $2.03$, ebullioscopic constant as $0.52$ in the above formula.
$ \Rightarrow \Delta {T_b} = 0.52 \times 2.03$
On simplifying we get
$ \Rightarrow \Delta {T_b} = 1$
So, the elevation in boiling point is ${1^ \circ }C$.
Now we will determine the boiling point of solution which is the sum of boiling point of water and elevation in boiling point.
The boiling point of water is ${100^ \circ }C$ and elevation in boiling point is ${1^ \circ }C$. So, the boiling point of solution is ${101^ \circ }C$

Hence, option $D$ is the correct option.

Note:
Always remember that the elevation in boiling point is defined as the colligative property which experiences an increase in the boiling point of the solvents when solutes are added into the solution. The colligative properties generally depend upon the number of particles.