Chemistry, 09.07.2019 21:00 genyjoannerubiera

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?

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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).

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

Learn more:

1. What is the equilibrium constant of pure water at ? link

2. Calculate the wavelength of the electron: link

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 from: Quest

The answer is 323 g.

Answer from: Quest

It has to be b i hope this

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Part a cyclohexane has a freezing point of 6.50 ∘c and a kf of 20.0 ∘c/m. what is the freezing point...

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