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.

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

he would use the colour black

the ratio of the conjugate base to the acid required to attain the desired ph.