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The molal elevation constant of water $ = 0.52K\,{m^{ - 1}}$ . The boiling point of $1.0$ molal aqueous $KCl$ solution (assuming complete dissociation of $KCl$ ) should be-
A. $100.52^\circ C$
B. $101.04^\circ C$
C. $99.48^\circ C$
D. $98.96^\circ C$

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Last updated date: 20th Jun 2024
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Answer
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Hint: There are four colligative properties of solution. Elevation in the boiling point of the solvent is one of them. The molal elevation constant is related to this colligative property and is denoted by ${K_b}$ . We also know that molality defines concentration of a solution.

Complete step by step answer:
We know that there are different ways in which concentration of a solution can be defined. Molality is one of them. It is defined as the number of moles of solute present per $1kg$ of solvent. Its unit is $mol/kg$ . Molality is independent of temperature since mass of any substance does not depend on temperature.
We know that vapour pressure of a pure volatile solvent decreases when a non-volatile solute is added to it. Due to this some properties of solvent are affected like-elevation of Boiling point of solvent, relative lowering of vapour pressure, depression in freezing point of solvent. All these properties depend on the number of particles of solute. And such properties are called colligative properties.
Elevation of boiling point: We know that vapour pressure of volatile solvent decreases when a non-volatile solute is added to it. The boiling point of volatile solvent is less than the solution having volatile solvent and non-volatile solute. This increase in boiling point due to the addition of solute is called as the elevation in the boiling point of solvent in solution. This can be understood from the example of sucrose solution. Water boils at $373K$ . Now when we add sucrose to water and make an aqueous solution, then the boiling point of this solution is $373.52K$ .
Let ${T_b}^\circ $ is the boiling point of pure solvent and ${T_b}$ be the boiling point of solution. Then the increase in boiling point can be given as-
$\Delta {T_b} = {T_b} - T_b^\circ $ where $\Delta {T_b}$ is the elevation of boiling point.
Experimentally it has been shown that for dilute solutions elevation is boiling point is directly proportional to the molal concentration of solute in a solution.
$\Delta {T_b}\alpha \,m$
$\Delta {T_b} = {K_b} \times m$
Here m is the molality or number of moles of solute per $kg$ of the solvent. ${K_b}$ is the molal elevation constant or also called Ebullioscopic constant. The unit of ${K_b}$ is $K\,kg\,mo{l^{ - 1}}\,or\,K\,{m^{ - 1}}$ .
 Now it is given that ${K_b}$ of water $ = 0.52K\,{m^{ - 1}}$ .
Molality of $KCl$ solution $m = 1.0\,mol\,k{g^{ - 1}}$.
When the solute dissolved in solvent is ionic, it dissociates into ions and the experimentally determined molar mass is always less than the true value. Van't Hoff introduced a factor $i$ , Van't Hoff factor to account for the dissociation.
$i = \dfrac{{Observed\,colligative\,property}}{{Calculated\,colligative\,property}}$
$ = \dfrac{{Total\,number\,of\,moles\,of\,particles\,after\,dissociation}}{{Total\,number\,of\,moles\,of\,particles\,before\,dissociation}}$
For $KCl$ , $i = 2$
Now the elevation in boiling point is given as-
$\Delta {T_b} = i \times {K_b} \times m$
Substituting the values we get elevation in boiling point-
$\Delta {T_b} = i \times {K_b} \times m$
$\Delta {T_b} = 2 \times 0.52 \times 1.0 = 1.04K$
Now the solvent is water and the solute is $KCl$ . The boiling point of solvent(water) is $373K$.
We have, $\Delta {T_b} = 1.04K\,,\,T_b^\circ = 373K$
$\Delta {T_b} = {T_b} - T_b^\circ $
So the boiling point of solution$({T_b})$ is-
${T_b} = \Delta {T_b} + T_b^\circ $
${T_b} = 1.04 + 373 = 374.04K$
So the boiling point of solution in $^\circ C$ will be-
$^\circ C = 374.04 - 273 = {101.04^
\circ }C$

So, the correct answer is Option B.

Note: The elevation in boiling point depends on the number of solute molecules and not on their nature. The other colligative properties like depression of freezing point, osmotic pressure and relative lowering of vapour pressure also depend on the number of solute molecules and not their nature.