Question

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.

Answer 1

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