Answer to Question #345558 in Inorganic Chemistry for Maryum

Question

What is the molality of para-dichlorobenzene in a solution prepared by dissolving 2.65g C6H4Cl2 in 50.0 mL of benzene (d=0.897g/mL)?

Answer 1

Solution:

The solute is para-dichlorobenzene (C₆H₄Cl₂)

The solvent is benzene (C₆H₆)

Density = Mass / Volume

Therefore,

Mass of C₆H₆ = Density × Volume = (0.897 g mL⁻¹) × (50.0 mL) = 44.85 g

Kilograms of C₆H₆ = (44.85 g) × (1 kg / 1000 g) = 0.04485 kg

Moles = Mass / Molar mass

The molar mass of para-dichlorobenzene (C₆H₄Cl₂) is 147 g mol⁻¹

Therefore,

Moles of C₆H₄Cl₂ = (2.65 g) / (147 g mol⁻¹) = 0.01803 mol

Molality = Moles of solute / Kilograms of solvent

Therefore,

Molality of C₆H₄Cl₂ solution = Moles of C₆H₄Cl₂ / Kilograms of C₆H₆

Molality of C₆H₄Cl₂ solution = (0.01803 mol) / (0.04485 kg) = 0.402 mol kg⁻¹ = 0.402 m

Molality of C₆H₄Cl₂ solution = 0.402 m

Answer: The molality of para-dichlorobenzene (C₆H₄Cl₂) solution is 0.402 m