Question

Choose the solvent below that would show the greatest boiling point elevation when used to make a 0.10 m nonelectrolyte solution. question 11 options: 1) ccl4, kb = 29.9°c/m 2) c6h6, kb = 5.12°c/m 3) ch3ch2och2ch3, kb = 1.79°c/m 4) ch3ch2oh, kb = 1.99°c/m 5) chcl3, kb = 4.70°c/m

Answer 1

The solvent that shows the greatest boiling point elevation to make a 0.10 m nonelectrolyte solution is $\boxed{{\text{1) CC}}{{\text{l}}_4}}$ .

Further Explanation:

Colligative properties:

The properties of a solution that depend only on the concentration of solute and not on the identity of the solute are known as colligative properties. The component present in smaller quantity is solute while that in larger quantity is known as a solvent. The solute gets itself dissolved in the solvent.

The colligative properties are as follows:

1. Relative lowering of vapor pressure.

2. Elevation in boiling point.

3. Depression in freezing point.

4. Osmotic pressure.

The temperature where the vapor pressure of the liquid and the atmospheric pressure becomes equal and the liquid changes into vapor is called the boiling point. It is a colligative property because it depends on the number of moles of solute particles present in the substance.

The formula to calculate the boiling point elevation is as follows:

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

Here,

$\Delta {{\text{T}}_{\text{b}}}$ is the elevation in boiling point.

${{\text{k}}_{\text{b}}}$ is the ebullioscopic constant.

m is the molality of the solution.

For a 0.10 m nonelectrolyte solution, the value of m is fixed in equation (1). So equation (1) modifies as follows:

$\Delta{{\text{T}}_{\text{b}}}\propto{{\text{k}}_{\text{b}}}$                                                                                …… (2)

More the value of ${{\text{k}}_{\text{b}}}$ , higher will be the elevation in the boiling point and vice-versa.

The value of ${{\text{k}}_{\text{b}}}$  for ${\text{CC}}{{\text{l}}_4}$  is $29.9\;^\circ{\text{C/m}}$ .

The value of ${{\text{k}}_{\text{b}}}$  for ${{\text{C}}_6}{{\text{H}}_6}$  is $5.12\;^\circ{\text{C/m}}$ .

The value of ${{\text{k}}_{\text{b}}}$  for ${\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{OC}}{{\text{H}}_2}{\text{C}}{{\text{H}}_3}$  is $1.79\;^\circ{\text{C/m}}$ .

The value of ${{\text{k}}_{\text{b}}}$  for ${\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{OH}}$  is $1.99\;^\circ{\text{C/m}}$ .

The value of ${{\text{k}}_{\text{b}}}$  for ${\text{CHC}}{{\text{l}}_3}$  is $4.70\;^\circ{\text{C/m}}$ .

${\text{CC}}{{\text{l}}_4}$  has the highest value of ${{\text{k}}_{\text{b}}}$  so in accordance with equation (2), it will have the greatest boiling point elevation.

Learn more:

1. What is the concentration of alcohol in terms of molarity? brainly.com/question/9013318

2. What is the molarity of the stock solution?: brainly.com/question/2814870

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Solutions

Keywords: kb, boiling point, CCl4, CHCl3, CH3CH2OCH2CH3, CH3CH2OH, C6H6, colligative properties, relative lowering of vapor pressure, elevation in boiling point, depression in freezing point, osmotic pressure, molality of solution, m, 0.10 m.

Answer 2

Answer is: 1) ccl4, kb = 29.9°c/m, carbon tetrachloride has the greatest boiling point elevation.

The boiling point elevation is directly proportional to the molality of the solution according to the equation: ΔTb = Kb · b.

ΔTb -  the boiling point elevation.
Kb - the ebullioscopic constant.
b - molality of the solution.
So the highest boiling poing elevation will be for solution with highest ebullioscopic constant because molality is the same.