Determine the freezing point of a solution that contains 78.8 g of naphthalene (C10H8, molar mass = 128.16 g/mol) dissolved in 722 mL of benzene (d = 0.877 g/mL). Pure benzene has a melting point of 5.50°C and a freezing point depression constant of 4.90°C/m.
delta Tf = Kf*molality
molality = number of moles of solute / mass of solvent in Kg
number of moles of solute = mass / molar mass
= 78.8 / 128.16
= 0.615 mol
volume of solvent = 722 mL
density = 0.877 g/mL
mass of solvent = density * volume
= 0.877 g/mL * 722 mL
= 633.2 g
= 0.6332 Kg
molality = number of moles of solute / mass of solvent
in Kg
= 0.615 mol / 0.6332 Kg
= 0.97 m
delta Tf = Kf*molality
= 4.9 * 0.97
= 4.76 oC
This is decraese in freezing point
so,
freezing point = 5.5 - 4.76 =0.74 oC
Answer: 0.74 oC
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