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Calculate the boiling point (in degrees C) of a solution made by dissolving 7 g of naphthalene {C10H8} in 14.4 g of benzene. The Kbp of the solvent is 2.53 K/m and the normal boiling point is 80.1 degrees C. Enter your answer using 2 decimal places.

User Uberma
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1 Answer

3 votes

Answer:

The boiling point = 89.69 °C

Step-by-step 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

User Mattforni
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