Increase in boiling point of a sucrose solution is $0.1K$, then what is increase in boiling point of the same concentration of $NaCl$ solution?
(1) $0.4K$
(2) $0.2K$
(3) $0.1K$
(4) $0.58K$

Hint: The elevation in boiling point of a solution varies directly with the concentration of the solution which indicates higher the concentration of solute in the solution, greater will be the elevation in the boiling point. Elevation in boiling point is a type of colligative property of matter i.e., it is dependent on the solute to solvent ratio but not on the solute’s identity.

Formula Used: $\Delta {T_b} = i{K_b}m$ where, $i$ is the Van’t Hoff factor and $m$ is the molality of the solute.

Complete Step by Step Solution:
A rise in the boiling point of the solution is observed when a non-volatile solute is added to it This increase in the boiling point of the solution is known as elevation in boiling point. This phenomenon is caused mainly because by adding solute, the concentration of the solution increases which requires a comparatively higher amount of temperature to boil.

The following equation for the elevation of boiling point allows it to be measured,
 $\Delta {{T}_{b}}=i{{K}_{b}}m$ [${{K}_{b}}=$Ebullioscopic constant]
i.e, elevation in boiling point$(\Delta {{T}_{b}})$ varies directly with molality $(m)$ and Van’t Hoff factor$(i)$. But as per the condition mentioned in the question, the concentration is the same for both the solutions which mean molality $(m)$ for both the solutions are the same. Then elevation in boiling point will vary according to the value of Van’t Hoff factor$(i)$.

Now the Van't Hoff factor$(i)$can be defined as the number of particles produced in solution per mole of the solute.

For sucrose${{C}_{12}}{{H}_{22}}{{O}_{11}}$, the value of Van’t Hoff factor$(i)$is $1$as sucrose does not break into particles in solution.
 $\therefore 0.1=1\times {{K}_{b}}\times m$ [Given $\Delta {{T}_{b}}=0.1K$for sucrose]
Or,$m{{K}_{b}}=0.1$ ……(1)
while the value of Van’t Hoff factor$(i)$ for$NaCl$ is$2$, since $NaCl$breaks into two particles or ions $(N{{a}^{+}}\And C{{l}^{-}})$
$NaCl\to N{{a}^{+}}+C{{l}^{-}}$

$\therefore \Delta {{T}_{b}}=2\times {{K}_{b}}\times m$
Or,$\Delta {{T}_{b}}=2\times 0.1$ [Putting the value of $m{{K}_{b}}$from the equation (1)]
Or,$\Delta {{T}_{b}}=0.2K$

Therefore, the increase in boiling point $NaCl$ will be twice that of sucrose i.e.,
Thus, the correct option is (B).

Note: It is important to note that the formula for elevation in boiling points holds true for non-volatile solvents only and not applicable to volatile solvents. Also, this formula becomes less precise when the concentration of the solute is very high.