Chapter 8, Problem 8.94EP

(a)

Interpretation Introduction

Interpretation:

The boiling point of solution containing $\text{1}\text{.0}\text{mol}$ of sucrose dissolved in $0.25\text{kg}$ solvent has to be identified.

Concept Introduction:

Boiling point is the temperature at which liquid turns into a gas. Example: boiling point of water is $\text{100°C}$. That is water changes from liquid phase to gas phase.

  ${\text{ΔT}}_{\text{b}}{\text{= T}}_{\text{b}}-{\text{T}}_{\text{b}}^{\text{°}}$

Where,

  ${\text{ΔT}}_{\text{b}}-$ Change in boiling point

  ${\text{T}}_{\text{b}}\text{ - }$ Boiling point of the solution

  ${\text{T}}_{\text{b}}^{\text{°}}\text{ - }$ Boiling point of pure solvent

Boiling point elevation $(\Delta T_b)$ is distinction between boiling point of the pure solvent ${\text{(T}}_{\text{b}}^{\text{°}}\text{)}$ and the boiling point of the solution ${\text{(T}}_{\text{b}}\text{)}$.

  ${\text{ΔT}}_{\text{b}}{\text{= iK}}_{\text{b}}\text{m}$

Where,

  ${\text{ΔT}}_{\text{b}}$ is the change in boiling point

  ${\text{K}}_{\text{b}}$ is the Molal boiling point constant

  M is the molality of the solution

  i is van’t Hoff factor

Note:

The boiling point of one kilogram of water will be increase by $\text{0}\text{.51°C}$ by the addition of one mole of solute particles.

Explanation of Solution

Given that $\text{1}\text{.0}\text{mol}$ of sucrose is added to make a solution in $0.25\text{kg}$ of water.

The addition of one mole of solute particles dissolved in one-kilogram of water will raise the boiling point of solution by $\text{0}\text{.51°C}$.

Therefore, the number of moles per kilogram in the solution has to be calculated and have to multiply by $\text{0}\text{.51°C}$.

One mole of dissolved sucrose will forms one mole of dissolved particles.

Boiling point increase can be calculated as follows:

  $\left[\frac{\text{1}\text{.0}\text{mol}sucrose}{0.25\text{kg}}\text{×}\left(0.51\text{°C}\right)\right]=2.04\text{°C}$

Boiling point calculation:

  $\text{100°C}\text{+}2.04\text{°C}\text{=}102.04\text{°C}$

(b)

Interpretation Introduction

Interpretation:

The boiling point of solution containing $\text{1}\text{.0}\text{mol}$ of sucrose dissolved in $0.500\text{kg}$ solvent has to be identified.

Concept Introduction:

Boiling point is the temperature at which liquid turns into a gas. Example: boiling point of water is $\text{100°C}$.  That is water changes from liquid phase to gas phase.

  ${\text{ΔT}}_{\text{b}}{\text{= T}}_{\text{b}}-{\text{T}}_{\text{b}}^{\text{°}}$

Where,

  ${\text{ΔT}}_{\text{b}}-$ Change in boiling point

  ${\text{T}}_{\text{b}}\text{ - }$ Boiling point of the solution

  ${\text{T}}_{\text{b}}^{\text{°}}\text{ - }$ Boiling point of pure solvent

Boiling point elevation $(\Delta T_b)$ is distinction between boiling point of the pure solvent ${\text{(T}}_{\text{b}}^{\text{°}}\text{)}$ and the boiling point of the solution ${\text{(T}}_{\text{b}}\text{)}$.

  ${\text{ΔT}}_{\text{b}}{\text{= iK}}_{\text{b}}\text{m}$

Where,

  ${\text{ΔT}}_{\text{b}}$ is the change in boiling point

  ${\text{K}}_{\text{b}}$ is the Molal boiling point constant

  M is the molality of the solution

  i is van’t Hoff factor

Note:

The boiling point of one kilogram of water will be increase by $\text{0}\text{.51°C}$ by the addition of one mole of solute particles.

Explanation of Solution

Given that $\text{1}\text{.0}\text{mol}$ of sucrose is added to make a solution in $0.500\text{kg}$ of water.

The addition of one mole of solute particles dissolved in one-kilogram of water will raise the boiling point of solution by $\text{0}\text{.51°C}$.

Therefore, the number of moles per kilogram in the solution has to be calculated and have to multiply by $\text{0}\text{.51°C}$.

One mole of dissolved sucrose will forms one mole of dissolved particles.

Boiling point increase can be calculated as follows:

  $\left[\frac{\text{1}\text{.0}\text{mol}\text{sucrose}}{\text{0}\text{.500}\text{kg}}\text{×}\left(\text{0}\text{.51°C}\right)\right]\text{=}\text{1}\text{.02°C}$

Boiling point calculation:

  $\text{100°C}\text{+}\text{1}\text{.02°C}\text{=}101.02\text{°C}$

(c)

Interpretation Introduction

Interpretation:

The boiling point of solution containing $\text{1}\text{.0}\text{mol}$ of sucrose dissolved in $1.00\text{kg}$ solvent has to be identified.

Concept Introduction:

Boiling point is the temperature at which liquid turns into a gas. Example: boiling point of water is $\text{100°C}$.  That is water changes from liquid phase to gas phase.

  ${\text{ΔT}}_{\text{b}}{\text{= T}}_{\text{b}}-{\text{T}}_{\text{b}}^{\text{°}}$

Where,

  ${\text{ΔT}}_{\text{b}}-$ Change in boiling point

  ${\text{T}}_{\text{b}}\text{ - }$ Boiling point of the solution

  ${\text{T}}_{\text{b}}^{\text{°}}\text{ - }$ Boiling point of pure solvent

Boiling point elevation $(\Delta T_b)$ is distinction between boiling point of the pure solvent ${\text{(T}}_{\text{b}}^{\text{°}}\text{)}$ and the boiling point of the solution ${\text{(T}}_{\text{b}}\text{)}$.

  ${\text{ΔT}}_{\text{b}}{\text{= iK}}_{\text{b}}\text{m}$

Where,

  ${\text{ΔT}}_{\text{b}}$ is the change in boiling point

  ${\text{K}}_{\text{b}}$ is the Molal boiling point constant

  M is the molality of the solution

  i is van’t Hoff factor

Note:

The boiling point of one kilogram of water will be increase by $\text{0}\text{.51°C}$ by the addition of one mole of solute particles.

Explanation of Solution

Given that $\text{1}\text{.0}\text{mol}$ of sucrose is added to make a solution in $1.00\text{kg}$ of water.

The addition of one mole of solute particles dissolved in one-kilogram of water will raise the boiling point of solution by $\text{0}\text{.51°C}$.

Therefore, the number of moles per kilogram in the solution has to be calculated and have to multiply by $\text{0}\text{.51°C}$.

One mole of dissolved sucrose will forms one mole of dissolved particles.

Boiling point increase can be calculated as follows:

  $\left[\frac{\text{1}\text{.0}\text{mol}\text{sucrose}}{\text{1}\text{.00}\text{kg}}\text{×}\left(\text{0}\text{.51°C}\right)\right]\text{=}\text{0}\text{.51°C}$

Boiling point calculation:

  $\text{100°C}\text{+}\text{0}\text{.51°C}\text{=}100.51\text{°C}$

(d)

Interpretation Introduction

Interpretation:

The boiling point of solution containing $\text{1}\text{.0}\text{mol}$ of sucrose dissolved in $2.00\text{kg}$ solvent has to be identified.

Concept Introduction:

Boiling point is the temperature at which liquid turns into a gas. Example: boiling point of water is $\text{100°C}$.  That is water changes from liquid phase to gas phase.

  ${\text{ΔT}}_{\text{b}}{\text{= T}}_{\text{b}}-{\text{T}}_{\text{b}}^{\text{°}}$

Where,

  ${\text{ΔT}}_{\text{b}}-$ Change in boiling point

  ${\text{T}}_{\text{b}}\text{ - }$ Boiling point of the solution

  ${\text{T}}_{\text{b}}^{\text{°}}\text{ - }$ Boiling point of pure solvent

Boiling point elevation $(\Delta T_b)$ is distinction between boiling point of the pure solvent ${\text{(T}}_{\text{b}}^{\text{°}}\text{)}$ and the boiling point of the solution ${\text{(T}}_{\text{b}}\text{)}$.

  ${\text{ΔT}}_{\text{b}}{\text{= iK}}_{\text{b}}\text{m}$

Where,

  ${\text{ΔT}}_{\text{b}}$ is the change in boiling point

  ${\text{K}}_{\text{b}}$ is the Molal boiling point constant

  M is the molality of the solution

  i is van’t Hoff factor

Note:

The boiling point of one kilogram of water will be increase by $\text{0}\text{.51°C}$ by the addition of one mole of solute particles.

Explanation of Solution

Given that $\text{1}\text{.0}\text{mol}$ of sucrose is added to make a solution in $2.00\text{kg}$ of water.

The addition of one mole of solute particles dissolved in one-kilogram of water will raise the boiling point of solution by $\text{0}\text{.51°C}$.

Therefore, the number of moles per kilogram in the solution has to be calculated and have to multiply by $\text{0}\text{.51°C}$.

One mole of dissolved sucrose will forms one mole of dissolved particles.

Boiling point increase can be calculated as follows:

  $\left[\frac{\text{1}\text{.0}\text{mol}\text{sucrose}}{\text{2}\text{.00}\text{kg}}\text{×}\left(\text{0}\text{.51°C}\right)\right]\text{=}\text{0}\text{.255°C}$

Boiling point calculation:

  $\text{100°C}\text{+}\text{0}\text{.255°C}\text{=}100.255\text{°C}$