Question

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)

Answer 1

The freezing point of a solution is -1.16°C

Calculation,

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

Δ$T_{f} = K_{f} .m$      ... (i)

• Δ$T_{f}$ is change in the freezing point of the solvent .

• $K_{f}$ 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  $K_{f}$ and m in equation (i) we get

Δ$T_{f} = K_{f} .m$  

Δ$T_{f}$= 5.12°C/m×1.3 m = 6.66°C

Δ$T_{f}$=  $T_{f}$ (solvent) - $T_{f}$ (solution)

$T_{f}$ (solution) = $T_{f}$ (solvent)  -Δ$T_{f}$= 5.5 °C - 6.66°C = -1.16°C

The freezing point of a is -1.16°C

To learn about freezing point

brainly.com/question/3121416

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