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What is the freezing point of a solution made from 22.0 g of octane (c8h18) dissolved in 148.0 g of benzene? (for benzene, freezing point = 5.50°c; kf = 5.12°c/m)

Respuesta :

The freezing point of a solution is -1.16°C

Calculation,

The equation which give relation between freezing point depletion and molality is,

Δ[tex]T_{f} = K_{f} .m[/tex]      ... (i)

  • Δ[tex]T_{f}[/tex] is change in the freezing point of the solvent .
  • [tex]K_{f}[/tex] of benzene = 5.12°C/m
  • m is molality of the solution

Moles of the solute = 22.0 g / 114.23 g/mole = 0.193 mole

Mass of the solvent in g = 148 g

Mass of the solvent in kg = 0.148 kg

Molality = 0.193 mole/ 0.148 kg = 1.3 m

Now, after putting the value of  [tex]K_{f}[/tex] and m in equation (i) we get

Δ[tex]T_{f} = K_{f} .m[/tex]  

Δ[tex]T_{f}[/tex]= 5.12°C/m×1.3 m = 6.66°C

Δ[tex]T_{f}[/tex]=  [tex]T_{f}[/tex] (solvent) - [tex]T_{f}[/tex] (solution)

[tex]T_{f}[/tex] (solution) = [tex]T_{f}[/tex] (solvent)  -Δ[tex]T_{f}[/tex]= 5.5 °C - 6.66°C = -1.16°C

The freezing point of a is -1.16°C

To learn about freezing point

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