Question

What would be the freezing point of a solution that has a molality of 1.468 m which was prepared by dissolving biphenyl (C12H10) into naphthalene? Relevant information is provided below. Provide your response to three digits after the decimal.

Answer 1

Answer:

Freezing point solution = 70.131 °C

Explanation:

Step 1: Data given

Molality = 1.468 molal

A solution is created by dissolving biphenyl (C12H10) into naphthalene

Biphenyl is a non-electrolyte

Freezing point of naphthalene = 80.26 °C

Step 2: Calculate the freezing point depression

ΔT = i*Kf*m

⇒with ΔT = the freezing point depression = TO BE DETERMINED

⇒with i = the van't Hoff factor of biphenyl = 1

⇒with Kf = the freezing point depression constant of naphthalene = 6.90 °C/m

⇒with m = the molality = 1.468 molal

ΔT = 1 * 6.90 °C/m * 1.468 °C

ΔT = 10.13 °C

Step 3: Calculate the freezing point of the solution

ΔT = 10.13 °C

Freezing point solution = freezing point naphthalene - 10.13 °C

Freezing point solution = 80.26 °C - 10.129 °C

Freezing point solution = 70.131 °C

Answer 2

The freezing point of the solution that has a molality of 1.468 m, prepared by dissolving biphenyl (C12H10) into naphthalene is 70.131 °C.

Depression at freezing point:

Given a solution with molality, m = 1.468

Let us represent the freezing point of naphthalene as Fp = 80.26 °C

Biphenyl is a non-electrolyte, by dissolving biphenyl (C12H10) into naphthalene, a freezing point depression is obtained as given below:

ΔT = i(Kf)m

where ΔT is the freezing point depression

i is van Hoff factor of biphenyl = 1

and Kf is the freezing point depression constant of naphthalene = 6.90 °C/m

ΔT = 1 × 6.90 × 1.468

ΔT = 10.13 °C

Now, the effective freezing point solution:

F = Fp - 10.13 °C

F = 80.26 °C - 10.129 °C

F = 70.131 °C

Learn more about freezing point depression:

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