Question

Part a cyclohexane has a freezing point of 6.50 ∘c and a kf of 20.0 ∘c/m. What is the freezing point of a solution made by dissolving 0.925 g of biphenyl (c12h10) in 25.0 g of cyclohexane?

Answer 1

The freezing point of the solution is heat loss during the given process is $\boxed{{\text{1}}{\text{.78}}^\circ{\text{C}}}$

Further explanation:

Freezing point:

It is defined as the temperature where both the solid and liquid phases are in equilibrium with each other. Here the vapor pressure of the substance in the liquid state becomes equal to the vapor pressure in a solid state.

The formula to calculate the change in freezing point is,

${\Delta }}{{\text{T}}_{\text{f}}}={{\text{k}}_{\text{f}}}{\text{m}}$                  ...... (1)

Here,

${\Delta }}{{\text{T}}_{\text{f}}}$ is the change in freezing point.

${{\text{k}}_{\text{f}}}$ is the freezing point depression constant.

m is the molality of the solution.

Molality is one of the concentration terms that is equal to the amount of solute divided by the mass of the solvent. It is represented by m and its unit is mol/kg.

The formula to calculate the molarity of the solution is,

${\text{Molality of the solution}}=\frac{{{\text{amount}}\;\left({{\text{mol}}}\right)\;{\text{of}}\;{\text{biphenyl}}}}{{\;{\text{mass}}\;\left({{\text{kg}}}\right)\;{\text{of}}\;{\text{cyclohexane}}}}$                  ……. (2)

The formula to calculate the moles of biphenyl is,

${\text{Moles of biphenyl}}=\frac{{{\text{Given mass of biphenyl}}}}{{{\text{Molar mass of biphenyl}}}}$              …… (3)

The given mass of biphenyl is 0.925 g.

The molar mass of biphenyl is 154.21 g/mol.

Substitute these values in equation (3).

$\begin{gathered}{\text{Moles of biphenyl}}=\left({{\text{0}}{\text{.925 g}}}\right)\left({\frac{{{\text{1 mol}}}}{{{\text{154}}{\text{.21 g}}}}}\right)\\={\text{0}}{\text{.0059 mol}}\\\end{gathered}$

The moles of biphenyl is 0.0059 mol.

The mass of cyclohexane is 25 g.

Substitute these values in equation (2).

$\begin{gathered}{\text{Molality}}=\frac{{\left({{\text{0}}{\text{.0059 mol}}}\right)}}{{\left({{\text{25 g}}}\right)}}\left({\frac{{{\text{1 g}}}}{{{\text{1}}{{\text{0}}^{{\text{-3}}}}\;{\text{kg}}}}}\right)\\=0.236\;{\text{m}}\\\end{gathered}$

The molarity is 0.236 m.

The value of ${{\text{k}}_{\text{f}}}$ is $20\;^\circ{\text{C/m}}$.

Substitute these values in equation (1).

$\begin{gathered}\Delta{{\text{T}}_{\text{f}}}=\left({\frac{{{\text{20}}\;{\text{^\circC}}}}{{{\text{1 m}}}}}\right)\left({{\text{0}}{\text{.0236 m}}}\right)\\=4.72\;{^\circ C}}\\\end{gathered}$

The formula to determine the change in freezing point is as follows:

${\Delta }}{{\text{T}}_{\text{f}}}={{\text{T}}_{\text{f}}}_{\left({{\text{solvent}}}\right)}-{{\text{T}}_{\text{f}}}_{\left({{\text{solution}}}\right)}$                      ...... (4)

Here,

${{\text{T}}_{\text{f}}}_{\left({{\text{solvent}}}\right)}$ is the temperature of the solvent.

${{\text{T}}_{\text{f}}}_{\left({{\text{solution}}}\right)}$ is the temperature of the solution.

Rearrange equation (4) to calculate the temperature of the solution.

${{\text{T}}_{\text{f}}}_{\left({{\text{solution}}}\right)}={{\text{T}}_{\text{f}}}_{\left({{\text{solvent}}}\right)}-{\Delta}}{{\text{T}}_{\text{f}}}$                              ……. (5)

Substitute $6.50\;^\circ{\text{C}}$ for ${{\text{T}}_{\text{f}}}_{\left({{\text{solvent}}}\right)}$ and $4.72\;^\circ{\text{C}}$ for ${\Delta }}{{\text{T}}_{\text{f}}}$ in equation (5).

$\begin{gathered}{{\text{T}}_{\text{f}}}_{\left({{\text{solution}}}\right)}=6.50\;^\circ{\text{C}}-4.72\;^\circ{\text{C}}\\=1.78\;^\circ{\text{C}}\\\end{gathered}$

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Colligative properties

Keywords: freezing point, temperature of solution, temperature of solvent, 1.78, molarity, biphenyl, cyclohexane, 0.236 m, m, mol/kg, solid, liquid.

Answer 2

The freezing point of the solution is 1.7 ∘C.

Recall that freezing point is a colligative property and the freezing point of a solution is less than the freezing point of the pure solvent.

Using the formula;

ΔT = K m i

ΔT = Freezing point depression

m = molaity of solution

i = Van't Hoff factor

From the question;

ΔT = ?

K = 20.0 ∘c/m

mass of solute =   0.925 g

Mass of solution = 25.0 g or 0.025 Kg

We now find the number of moles of solute; 0.925 g/154 g/mol

= 0.006 moles

Molality of solution = 0.006 moles/0.025 Kg = 0.24 m

So;

ΔT = 20.0 ∘C/m × 0.24 m × 1

i = 1 because the compound is molecular

ΔT = 4.8∘C

But ΔT = Freezing point of pure solvent - Freezing point of solution

Freezing point of pure solvent = 6.50 ∘C

4.8∘C = 6.50 ∘c - Freezing point of solution

Freezing point of solution =  6.50 ∘C - 4.8∘C

Freezing point of solution = 1.7 ∘C

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