Question

For the solution resulting from dissolved 0.32 g of naphthalene (C10H8) in 25 g of benzene (C6H6) at temperature of 26.1°C, calculate the vapor pressure lowering, the boiling point elevation, and the freezing point depression. The vapor pressure of benzene at the temperature of the experiment is 100 torr. (Kf of benzene = 2.67 °C/m, Kb of benzene = 5.12 °C/m)

Answer 1

Answer:

See explanation

Explanation:

Number of moles of naphthalene = 0.32g/128.1705 g/mol = 0.0025 moles

Molality = number of moles/ mass of Solvent in kilograms

Molality = 0.0025/0.025 Kg

Morality = 0.1 m

But

∆T= K × i × m

Where ∆T = boiling point elevation

i= number of particles (this is equal to 1 because naphthalene is molecular and not ionic)

m= molality of naphthalene = 0.1 m

K= boiling point elevation constant = 5.12 °C/m

∆T= 5.12 °C/m ×0.1 = 0.512°C

For freezing point depression

∆T= K× i × m

Where ∆T= freezing point depression

i= number of particles (this is equal to 1 because naphthalene is molecular and not ionic)

m= molality of naphthalene = 0.1 m

K= freezing point depression constant = 2.67 °C/m

∆T= 2.67 °C/m ×0.1 = 0.267°C

From Raoult's law;

∆P = XBPA°

Where;

∆P = vapour pressure lowering

XB = mole fraction of solute

PA° = vapour pressure of pure solvent

Number of moles of solvent = mass/molar mass = 25g/ 78 g/mol= 0.3205 moles

Total number of moles = number of moles of solute + number of moles of solvent = 0.0025 moles + 0.3205 moles = 0.323 moles

Mole fraction of solute = 0.0025 moles/0.323 moles = 0.0077

Vapour pressure of benzene = 100 torr

Therefore;

∆P = 0.0077 × 100torr = 0.77 torr

Hence;

∆P = 0.77 torr