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Calculate the freezing point and boiling point of a solution containing 12.4 g of naphthalene (C10H8) in 101.0 mL of benzene. Benzene has a density of 0.877 g/cm3.

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Answer:

boiling point = 82.86 °C

Freezing point = -0.08 °C

Explanation:

Step 1:

Mass of naphtalene = 12.4 grams

Molar mass naphtalene = 128.17 g/mol

Volume of benzene = 101.0 mL

Density benzene = 0.877 g/cm3 = 0.877 g/mL

Kf(benzene)= 5.12°C/m

Freezing Point  benzene = 5.5 °C

Kb(benzene)=2.53°C/m

Boiling Point benzene = 80.1 °C

Step 2: Calculate mass benzene

Mass Benzene = density * volume

Mass benzene = 0.877 g/mL *101.0 mL

Mass benzene = 88.577 grams

Step 3: Calculate moles naphthalene

Moles naphthalene = 12.4 grams / 128.17 g/mol

Moles naphthalene = 0.0967 moles

Step 4: Calculate molality

molality = moles naphthalene / mass benzene

Molality = 0.0967 moles / 0.088577 kg

Molality = 1.09 molal

Step 5: Calculate boiling point of solution

ΔT = i*Kb*m

⇒ with i = Van't Hoff factor of naphthalene = 1

⇒ with Kb(benzene) = 2.53°C/m

⇒ with m = molality = 1.09 molal

ΔT = 1*2.53 °C/m * 1.09 m

ΔT = 2.76 °C

boiling point = 80.1 + 2.76 = 82.86 °C

Step 6: Calculate freezing point of solution

ΔT = i*Kb*m

⇒ with i = Van't Hoff factor of naphthalene = 1

⇒ with Kf(benzene) = 5.12 °C/m

⇒ with m = molality = 1.09 molal

ΔT = 1*5.12 °C/m * 1.09 m

ΔT = 5.58 °C

Freezing point = 5.5 °C - 5.58 °C = -0.08 °C

pstnonsonjoku

The boiling point of the solution is 82.86 °C while the freezing point of the solution is -0.08 °C.

We know that;

  • Mass of naphtalene = 12.4 grams
  • Molar mass naphtalene = 128.17 g/mol
  • Volume of benzene = 101.0 mL
  • Density benzene = 0.877 g/cm3 = 0.877 g/mL
  • Kf(benzene)= 5.12°C/m
  • Freezing Point  benzene = 5.5 °C
  • Kb(benzene)=2.53°C/m
  • Boiling Point benzene = 80.1 °C

To obtain the mass of  benzene

Mass Benzene = density * volume

= 0.877 g/mL *101.0 mL

= 88.577 grams

To obtain the number of moles of naphthalene

Number of moles = 12.4 grams / 128.17 g/mol

= 0.0967 moles

The molality is obtained as follows;

molality = moles naphthalene / mass benzene

= 0.0967 moles / 0.088577 kg

= 1.09 molal

The boiling point of solution is obtained as follows;

ΔT = i*Kb*m

  • i = Van't Hoff factor
  • Kb = boiling point constant
  • m = molality

ΔT = 1*2.53 °C/m * 1.09 m

ΔT = 2.76 °C

boiling point solution = boiling point of pure solvent + ΔT

boiling point of solution = 80.1 + 2.76 = 82.86 °C

To obtain the  freezing point of solution;

ΔT = i*K*m

ΔT = 1*5.12 °C/m * 1.09 m

ΔT = 5.58 °C

Freezing point of solution = 5.5 °C - 5.58 °C = -0.08 °C

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