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
