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This question deals with colligative properties. Let us first look into colligative properties and the factors on which they depend.

Colligative properties of solutions arise due to the presence of non-volatile solute in a solution. They depend on the number of the solute particles and not on the nature of the chemical species present in a solution. When a non-volatile solute is added to a solvent, it displaces some solvent molecules at the surface of the solution thereby reducing vapour pressure of the solvent which is called the relative lowering of vapour pressure.

A liquid will start to boil at a particular temperature if its vapour pressure becomes equal to the atmospheric pressure at that temperature. Since the vapour pressure of a liquid decreases on the addition of a non-volatile solute, therefore its temperature has to be raised so that more vapours form resulting in an increase in the vapour pressure and when the vapour pressure becomes equal to the atmospheric pressure, the liquid will start to boil. Hence the elevation in boiling point of a liquid is observed on the addition of a non-volatile solute.

We can apply the calculations for colligative properties only for the dilute solutions since their behaviour is very close to that of an ideal solution.

The formula for the elevation in boiling point is:

$\triangle { T }_{ b }={ K }_{ b }m$

Where $\triangle { T }_{ b }$ is the difference between the boiling point of the solution and the pure solvent, ${ K }_{ b }$ is the molal elevation constant and m is the molality of the solution.

The molar mass of naphthalene (${ C }_{ 10 }{ H }_{ 8 }$) is 128 g/mol.

Therefore the number of moles of naphthalene will be =$\cfrac { 5.00g }{ 128\quad g/mol } =0.039\quad mol$

The mass of the solvent benzene =100 g which is equal to 0.100 Kg. Hence the molality of the solution will be:

Molality =$\cfrac { 0.039\quad mol }{ 0.100\quad Kg } =0.39\quad mol/Kg$

Using the equation: $\triangle { T }_{ b }={ K }_{ b }m$, we will find the elevation in the boiling point:

$\Delta { T }_{ b }=2.53^{ o }{ C{ mol }^{ -1 }\times 0.39\quad mol=0.987^{ o }{ C\simeq 1^{ o }{ C } } }$

Since the boiling point of pure benzene is 80$^{ o }{ C}$, therefore the boiling point of the solution will be=$80^{ o }{ C }+1^{ o }{ C }=81^{ o }{ C }$