Question

Calculate the freezing point and boiling point of a solution containing 14.2 gg of naphthalene (C10H8)(C10H8) in 111.0 mLmL of benzene. Benzene has a density of 0.877 g/cm3g/cm3.

Answer 1

Answer:

The freezing point of solution = -0.34 °C

The boiling point of solution = 82.98 °C

Explanation:

Step 1: Data given

Mass of naphthalene = 14.2 grams

Molar mass naphthalne = 128.17 g/mol

Volume of benzene = 111.0 mL

Density of benzene = 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 of benzene

Mass benzene = density * volume

Mass benzene = 0.877 g/mL * 111.0 mL

Mass benzene = 97.3 grams

Step 3: Calculate moles naphthalene

Moles naphthalene = mass naphthalene / molar mass napthalene

Moles napthalene = 14.2 grams / 128.17 g/mol

Moles naphthalene = 0.111 moles

Step 4: Calculate molality

Molality = moles naphthalene / mass benzene

Molality = 0.111 moles / 0.0973 kg

Molality = 1.14 molal

Step 5: Calculate freezing point  of a solution

ΔT = i*kf*m

ΔT = 1 * 5.12 °C/m * 1.14 m

ΔT = 5.84 °C

ΔT = T(pure solvent) − T(solution)

The freezing point of solution = T pure -  ΔT

5.5 - 5.84 = -0.34 °C

The freezing point is -0.34 °C

Step 6: Calculate boiling point  of a solution

ΔT = i*kb*m

ΔT = 1 * 2.53 °C/m * 1.14 m

ΔT = 2.88 °C

ΔT = Tb (solution) - Tb (pure solvent)

The boiling point of solution = T pure +  ΔT

The boiling point of solution = 80.1 °C + 2.88

The boiling point of solution = 82.98 °C