Determine the boiling point of a solution that contains 150.0 g of naphthalene (C10H8, molar mass = 128.16 g/mol) dissolved in 950 mL of benzene (d = 0.877 g/mL). Pure benzene has a boiling point of 80.1°C and a boiling point elevation constant of 2.53°C/m.

Boiling T° of solution  → 83.6°C

Explanation:

To solve this, we apply Elevation of boiling point, property

ΔT = Kb . m . i

As we talk about organic solute, i = 1. No ions are formed.

m = molality (moles of solute in 1kg of solvent)

We determine mass of solvent by density

D = m /V so D . V = m

950 mL . 0.877 g/mL = 833.15 g

We convert to kg → 833.15 g . 1 kg/ 1000g = 0.833 kg

Moles of solute (naphtalene): 150 g . 1 mol/ 128.16g = 1.17 mol

m = 1.17mol / 0.833 kg = 1.41 mol/kg

We replace data:

Boiling T° of solution - 80.1°C = 2.53°C/m . 1.41 m . 1

Boiling T° of solution = 2.53°C/m . 1.41 m . 1  +  80.1°C → 83.6°C