Question

Calculate the freezing point of a solution containing 1.25g of benzene in 100g of chloroform

Answer 1

Freezing point of solution containing 1.25 g of benzene in 100 g of chloroform is $\boxed{ - 64.25{\text{ }}^\circ {\text{C}}}$.

Further Explanation:

The properties that depend only on concentration of solute particles, not on their identities are called colligative properties. Four colligative properties are enlisted below.

• Relative lowering of vapor pressure
• Elevation in boiling point
• Depression in freezing point
• Osmotic pressure

The expression for freezing point depression is,

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

Here,

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

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

m is molality of solution.

The formula to calculate moles of benzene is as follows:

${\text{Moles of benzene}} = \dfrac{{{\text{Mass of benzene}}}}{{{\text{Molar mass of benzene}}}}$             …… (2)

Substitute 1.25 g for mass of benzene and 78.11 g/mol for molar mass of benzene in equation (2).

$\begin{aligned}{\text{Moles of benzene}}&= \frac{{{\text{1}}{\text{.25 g}}}}{{{\text{78}}{\text{.11 g/mol}}}}\\&= 0.016{\text{ mol}} \\\end{aligned}$  

The molality of benzene solution is calculated as follows:

 $\begin{aligned}{\text{m}}&= \left( {\frac{{0.016{\text{ mol}}}}{{100.0{\text{ g}}}}} \right)\left( {\frac{{1{\text{ g}}}}{{{{10}^{ - 3}}{\text{ kg}}}}} \right)\\&= 0.16{\text{ m}} \\ \end{aligned}$

Substitute 0.16 m for m and $4.68{\text{ }}^\circ {\text{C/m}}$ for ${{\text{k}}_{\text{f}}}$ in equation (1).

$\begin{aligned}\Delta {{\text{T}}_{\text{f}}} &= \left( {4.68{\text{ }}^\circ {\text{C/m}}} \right)\left( {{\text{0}}{\text{.16 m}}} \right) \\&= 0.7488{\text{ }}^\circ {\text{C}} \\\end{aligned}$  

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

 ${{\Delta }}{{\text{T}}_{\text{f}}} = {{\text{T}}_{\text{f}}}_{\left( {{\text{chloroform}}} \right)} - {{\text{T}}_{\text{f}}}_{\left( {{\text{solution}}} \right)}$                                                                                            ...... (3)

Where,

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

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

Rearrange equation (3) to calculate the freezing point of solution.

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

Substitute $- 63.5{\text{ }}^\circ {\text{C}}$ for ${{\text{T}}_{\text{f}}}_{\left( {{\text{chlorofrom}}} \right)}$ and $0.7488{\text{ }}^\circ {\text{C}}$ for $\Delta {{\text{T}}_{\text{f}}}$ in equation (4).

$\begin{aligned}{{\text{T}}_{\text{f}}}_{\left( {{\text{solution}}} \right)} &= - 63.5{\text{ }}^\circ {\text{C}} - 0.7488{\text{ }}^\circ {\text{C}} \\&= - 64.25{\text{}}^\circ {\text{C}} \\\end{aligned}$  

Learn more:

1. Choose the solvent that would produce the greatest boiling point elevation: brainly.com/question/8600416
2. What is the molarity of the stock solution? brainly.com/question/2814870

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Colligative properties

Keywords: colligative properties, freezing point, chloroform, benzene, mass, molar mass, 1.25 g, 100 g, 78.11 g/mol, 0.16 m, 0.016 mol.

Answer 2

Answer:

The freezing point of chloroform (CHCl₃) in presence of benzene = - 64.252 °C.

Explanation:

• We can solve this problem using the relation:

ΔTf = Kf.m,

Where, ΔTf is the elevation of freezing point,

Kf is the freezing point depression constant (Kf of chloroform = 4.7 °C/m),

m is the molality of the solution.

• We can get the molality of benzene from the relation:

m = (mass / molar mass) solute (benzene) x (1000 / mass of solvent (CHCl₃))

m = (1.25  g / 78.11 g/mol) x (1000 / 100.0 g) = 0.160 m.

ΔTf = Kf.m = (4.70 °C/m) (0.160 m) = 0.752 °C.

∴ The freezing point in presence of benzene = The freezing point of pure CHCl₃ - ΔTf = - 63.50 °C – 0.752 °C = - 64.252 °C.