Calculate the boiling point of a solution prepared by dissolving 70.0 g of naphthalene, C10H8 (a nonvolatile nonelectrolyte), in 220.0 g of benzene, C6H6. The Kb for benzene = 2.53oC/m. The boiling point of pure benzene is 80.1oC.. . Ans: 86.4 degrees Celsius. I did this so far. 1)70G C10H8(1mol c10H8/128gC10H8)= .546mol C10H8. 2)(.546mol c10h8)(.220kg benzene)=2.482m. 3) (2.482)(2.53 C/m)=6.289. I'm not sure if I am starting this off right, can anyone help me get the correct answer? Please ans ty!

The boiling point of the solution is the sum of the boiling point rise and the boiling point of the solvent. Boiling point rise is Kb multiplied by the molality of the solution. ΔTb hence is 2.53 C/m *[70g/128 g/mol/0.220 kg], equal to 6.289 C. Hence the boiling point of the solution is 80.1 + 6.289 C equal to 86.39 or 86.4 C.

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Tringa0 Tringa0 · Answer 2 · 2019-10-07T07:37:30+02:00

Answer:

86.4°C is the boiling point of a solution.

Explanation:

[tex]\Delta T_b=T_b-T[/tex]

[tex]\Delta T_b=K_b\times m[/tex]

[tex]\Delta T_b=iK_b\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in Kg}}[/tex]

where,

Mass of benzene = 220.0 g = 0.220 kg

Mass of solute or naphthalene = 70.0 g

[tex]\Delta T_f[/tex] =Elevation in boiling point

[tex]K_b[/tex] = Boiling point constant of solvent = 2.53 °C/m(benzene)

1 = van't Hoff factor (organic solute)

m = molality

[tex]\Delta T_b=1\times 2.53^oC/m\times \frac{70.0 g}{128 g/mol\times 0.220 kg}[/tex]

[tex]\Delta T_b=6.27^oC[/tex]

[tex]\Delta T_b=T_b-T[/tex]

[tex]6.27^oC=T-80.1^oC[/tex]

[tex]T_b=6.29^oC+80.1^oC=86.39^oC\approx 86.4^oC[/tex]

86.4°C is the boiling point of a solution.