Calculate the boiling point of a solution prepared by adding 11.5 g naphthalene (C10H8) to 250.0 g of benzene. Naphthalene is a non-electrolyte solute, and benzene is an organic solvent that exhibits a boiling point of 80.10 oC, and has a kb

81.0°C

Explanation:

Kb benzene = 2.5°C/m

The addition of a solute to a pure solvent produce an elevation in boiling point regarding to boiling point of pure solvent. This phenomenon follows the equation:

ΔT = Kb×m×i

where ΔT represents the increasing in boiling point, Kb is the elevation boiling point constant of the solvent (2.5°C/m for benzene), m is molality of solution (Moles solute / kg solvent) and i is Van't Hoff factor (1 for a non-electrolyte solute as naphthalene).

Moles of 11.5g of naphthalene (Molar mass: 128.17g/mol) are:

11.5g × (1mol / 128.17g) = 0.0897 moles of naphthalene in 250.0g = 0.250kg of benzene.

Molality is:

0.0897 moles of naphthalene / 0.250kg of benzene = 0.359m

Replacing in the equation:

ΔT = Kb×m×i

ΔT = 2.5°C/m×0.359m×1

ΔT = 0.90°C

That means the solution prepared has an elevation in boiling point of 0.90°C. As boiling point of pure benzene is 80.10°C, boiling point of the solution is:

80.10°C + 0.90°C =

i think it is d. e because the next one is igneous.

the answer is a and have a great day