Chapter 13, Problem 13.71QP

(a)

Interpretation Introduction

Interpretation:

For following solutions normal freezing point and boiling point to be calculated.

1. a) $\text{21}\text{.2g NaCl in 135 mL}$ water
2. b)

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 ${\text{(T}}_{\text{b}}\text{)}$.
•  

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

Where,

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

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

m- molality of the solution

• Freezing point is the temperature at which liquid turns into solid.
• Freezing point depression ${\text{(ΔT}}_{\text{f}}\text{)}$ is distinction between freezing point of the pure solvent ${\text{(T}}_{\text{f}}^{\text{°}}\text{)}$ and the freezing point of the solution ${\text{(T}}_{\text{f}}\text{)}$.
•  

  ${\text{ΔT}}_{\text{f}}{\text{= K}}_{\text{f}}\text{m}$

Where,

${\text{ΔT}}_{\text{f}}-$ Change in freezing point

${\text{K}}_{\text{f}}-$ Molal freezing point constant

m- Molality of the solution

Answer to Problem 13.71QP

Freezing point of $\text{NaCl}$  solution = $\text{-10}\text{.0°C}$

Boiling point of $\text{NaCl}$  solution = $\text{102}\text{.8°C}$

Explanation of Solution

To record the given data

Amount of water in which NaCl present = $\text{135mL}$

Amount of Sodium chloride = $\text{21}\text{.2g}$

Amount of water and Sodium chloride are recorded as shown.

Sodium chloride is strong electrolyte. The concentration of the particles is double this solution. Note that because the density of water is 1g/mL and mass of water 135g.

To calculate mole of Sodium chloride

$\text{21}\text{.2g ×}\frac{\text{1mol}}{\text{58}\text{.44g}}\text{= 0}\text{.363mol}$

By plugging in the values of amount and molecular mass of Sodium chloride, a mole of Sodium chloride has calculated.

Calculation of molality of the solution

$\text{molality =}\frac{\text{0}\text{.363 mol}}{\text{0}\text{.135kg water}}\text{= 2}\text{.70m}$

By plugging in the value of moles of the solute in kilogram water, molality of the solution has calculated.

Finding change in freezing point and boiling point of the solution

i=2

${\text{ΔT}}_{\text{f }}{\text{= iK}}_{\text{f}}\text{m = 2(1}\text{.86°C/m)(2}\text{.70m) =10}\text{.0°C}$

${\text{ΔT}}_{\text{b }}{\text{= iK}}_{\text{b}}\text{m = 2(0}\text{.52°C/m)(2}\text{.70m) =2}\text{.8°C}$

New boiling point = $\text{100 + 2}\text{.8°C=102}\text{.8}°\text{C}$

By plugging in the value of molal freezing point constant and molality of the solution, change in freezing point and boiling point of the solution has calculated.

Conclusion

For given solutions normal freezing point and boiling point has been calculated.

Freezing point of $\text{NaCl}$  solution = $\text{-10}\text{.0°C}$

Boiling point of $\text{NaCl}$  solution = $\text{102}\text{.8°C}$

(b)

Interpretation Introduction

Interpretation:

For following solutions normal freezing point and boiling point to be calculated.

(b)$\text{15}\text{.4g of Urea in 66}\text{.7mL of water}$

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 ${\text{(T}}_{\text{b}}\text{)}$.

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

Where,

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

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

m- molality of the solution

• Freezing point is the temperature at which liquid turns into solid.
• Freezing point depression ${\text{(ΔT}}_{\text{f}}\text{)}$ is distinction between freezing point of the pure solvent ${\text{(T}}_{\text{f}}^{\text{°}}\text{)}$ and the freezing point of the solution ${\text{(T}}_{\text{f}}\text{)}$.

${\text{ΔT}}_{\text{f}}{\text{= K}}_{\text{f}}\text{m}$

Where,

${\text{ΔT}}_{\text{f}}-$ Change in freezing point

${\text{K}}_{\text{f}}-$ Molal freezing point constant

m- Molality of the solution

Answer to Problem 13.71QP

Freezing point of Urea solution = $\text{-7}\text{.14°C}$

Boiling point of Urea solution = $\text{102}\text{.0°C}$

Explanation of Solution

Given data

Amount of water in which Urea present = $\text{66}\text{.7 mL}$

Amount of Urea = $\text{15}\text{.4g}$

Amount of water and urea are recorded as shown.

Calculation of mole of Urea

Urea is non- electrolyte. The concentration of the particles is just equal to the concentration of urea.

Molecular mass of urea = 60.06g

$\text{15}\text{.4g ×}\frac{\text{1mol}}{\text{60}\text{.06g}}\text{= 0}\text{.256 mol}$

By plugging in the values of amount and molecular mass of Urea, a mole of Urea has calculated.

Calculate\ion of molality of the solution

$\text{molality =}\frac{\text{0}\text{.256 mol}}{\text{0}\text{.0667kg water}}\text{= 3}\text{.84 m}$

By plugging in the value of moles of the solute in kilogram water, molality of the solution has calculated.

Finding change in freezing point and boiling point of the solution

i=1

${\text{ΔT}}_{\text{f }}{\text{= iK}}_{\text{f}}\text{m = 1(1}\text{.86°C/m)(3}\text{.84m) =7}\text{.14°C}$

${\text{ΔT}}_{\text{b }}{\text{= iK}}_{\text{b}}\text{m =1(0}\text{.52°C/m)(3}\text{.84m) =2}\text{.0°C}$

 boiling point = $\text{100 + 2}\text{.8°C=102}\text{.0}°\text{C}$

By plugging in the value of molal freezing point constant and molality of the solution, change in freezing point and boiling point of the solution has calculated.

Conclusion

For given solutions normal freezing point and boiling point has been calculated.

Freezing point of Urea solution was calculated as $\text{-7}\text{.14°C}$

Boiling point of Urea solution was calculated as $\text{102}\text{.0°C}$