Question

What would be the freezing point of a solution that has a molality of 1.324 m which was prepared by dissolving biphenyl (C12H10) into naphthalene

Answer 1

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

A solution is prepared by dissolving biphenyl into naphthalene. We can calculate the freezing point depression (ΔT) for naphthalene using the following expression.

$\Delta T = i \times Kf \times m = 1 \times 6.90 \°C/m \times 1.324m = 9.14 \°C$

where,

• i: van 't Hoff factor (1 for non-electrolytes)
• Kf: cryoscopic constant
• m: molality

The normal freezing point of naphthalene is 80.26 °C. The freezing point of the solution is:

$T = 80.26 \° C - 9.14 \° C = 71.12 \° C$

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

Learn more: brainly.com/question/2292439