Answer:
The boiling point = 89.69 °C
Explanation:
Step 1: Data given
Mass of naphthalane = 7.0 grams
Mass of benzene = 14.4 grams
The Kbp of the solvent = 2.53 K/m
The normal boiling point is 80.1°C
Naphthalene, C10H8 , is a non-electrolyte, which means that the van't Hoff factor for this solution will be 1
Step 2: Calculate moles naphthalene
Moles naphthalene = mass / molar mass
Moles naphthalene = 7.0 grams / 128.17 g/mol
Moles naphthalene = 0.0546 moles
Step 3: Calculate molality
Molality = moles naphthalene / mass water
Molality = 0.0546 moles / 0.0144 kg
Molality = 3.79 molal
Step 4:
ΔT = i*Kb*m
ΔT = 1*2.53 K/m * 3.79 molal
ΔT = 9.59 °C
The boiling point = 80.1 °C + 9.59 °C = 89.69 °C
