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