Respuesta :
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.
Freezing point of solution containing 1.25 g of benzene in 100 g of chloroform is [tex]\boxed{ - 64.25{\text{ }}^\circ {\text{C}}}[/tex].
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,
[tex]\Delta {{\text{T}}_{\text{f}}} = {{\text{k}}_{\text{f}}}{\text{m}}[/tex] …… (1)
Here,
[tex]\Delta {{\text{T}}_{\text{f}}}[/tex] is depression in freezing point.
[tex]{{\text{k}}_{\text{f}}}[/tex] is molal freezing point constant.
m is molality of solution.
The formula to calculate moles of benzene is as follows:
[tex]{\text{Moles of benzene}} = \dfrac{{{\text{Mass of benzene}}}}{{{\text{Molar mass of benzene}}}}[/tex] …… (2)
Substitute 1.25 g for mass of benzene and 78.11 g/mol for molar mass of benzene in equation (2).
[tex]\begin{aligned}{\text{Moles of benzene}}&= \frac{{{\text{1}}{\text{.25 g}}}}{{{\text{78}}{\text{.11 g/mol}}}}\\&= 0.016{\text{ mol}} \\\end{aligned}[/tex]
The molality of benzene solution is calculated as follows:
[tex]\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}[/tex]
Substitute 0.16 m for m and [tex]4.68{\text{ }}^\circ {\text{C/m}}[/tex] for [tex]{{\text{k}}_{\text{f}}}[/tex] in equation (1).
[tex]\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}[/tex]
The formula to calculate the change in freezing point is as follows:
[tex]{{\Delta }}{{\text{T}}_{\text{f}}} = {{\text{T}}_{\text{f}}}_{\left( {{\text{chloroform}}} \right)} - {{\text{T}}_{\text{f}}}_{\left( {{\text{solution}}} \right)}[/tex] ...... (3)
Where,
[tex]{{\text{T}}_{\text{f}}}_{\left( {{\text{chloroform}}} \right)}[/tex] is temperature of pure chloroform.
[tex]{{\text{T}}_{\text{f}}}_{\left( {{\text{solution}}} \right)}[/tex] is freezing point of solution.
Rearrange equation (3) to calculate the freezing point of solution.
[tex]{{\text{T}}_{\text{f}}}_{\left( {{\text{solution}}} \right)} = {{\Delta }}{{\text{T}}_{\text{f}}} - {{\text{T}}_{\text{f}}}_{\left( {{\text{chloroform}}} \right)}[/tex] ……. (4)
Substitute [tex]- 63.5{\text{ }}^\circ {\text{C}}[/tex] for [tex]{{\text{T}}_{\text{f}}}_{\left( {{\text{chlorofrom}}} \right)}[/tex] and [tex]0.7488{\text{ }}^\circ {\text{C}}[/tex] for [tex]\Delta {{\text{T}}_{\text{f}}}[/tex] in equation (4).
[tex]\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}[/tex]
Learn more:
- Choose the solvent that would produce the greatest boiling point elevation: https://brainly.com/question/8600416
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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.
