Chapter 13.5, Problem 13.6WE

Ethylene glycol [CH₂(OH)CH₂(OH)] is a common automobile antifreeze. It is water soluble and fairly nonvolatile (b.p. 197°C). Calculate (a) the freezing point and (b) the boiling point of a solution containing 685 g of ethylene glycol in 2075 g of water.

Interpretation Introduction

Interpretation:

Freezing point and boiling point of the solution containing 685g of ethylene glycol has to determine.

Concept introduction:

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

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

Answer to Problem 13.6WE

Freezing point of the solution = $\text{-9}\text{.89°C}$

Explanation of Solution

Given

Amount of Ethylene glycol in water = $\text{625g}$

Amount of water in which Ethylene glycol dissolved= $\text{2075g}$

Boiling point of Ethylene glycol = $\text{197°C}$

Conversion of grams of ethylene glycol to moles

$\frac{{\text{685g C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}}{\text{62}\text{.07g/mol}}\text{= 11}\text{.04 mol}$

By plugging in the value of mass and molar mass of ethylene glycol, the number of moles of ethylene glycol has calculated.

Calculation of molality of the solution

$\frac{\text{11}{\text{.04mol C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}}{\text{2}\text{.075kg water}}\text{= 5}\text{.32 m}$

By plugging in the value of moles of ethylene glycol and amount of water, the molality of the solution has calculated.

Calculation of molality of glucose solution

$\text{1062 g solution -170}\text{.1g glucose = 892g water = 0}\text{.892kg water}$

$\frac{\text{0}\text{.9440mol}}{\text{0}\text{.892kg}}\text{= 1}\text{.06m}$

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

Calculation of freezing point of the solution

${\text{ΔT}}_{\text{f}}{\text{ = K}}_{\text{f}}\text{m =(1}\text{.86°C/m)(5}\text{.32m) = 9}\text{.89°C}$

The freezing point of the solution is (0- $\text{9}\text{.89°C)}$= $\text{-9}\text{.89°C}$

By plugging in the value of freezing point and molal freezing point depression constant for ethylene glycol, freezing point of the solution has calculated.

Conclusion

Freezing point of the solution has calculated as $\text{-9}\text{.89°C}\text{.}$

Interpretation Introduction

Interpretation:

Freezing point and boiling point of the solution containing 685g of ethylene glycol has to determine.

Concept introduction:

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

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

Answer to Problem 13.6WE

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

Explanation of Solution

Given

Amount of Ethylene glycol in water = $\text{625g}$

Amount of water in which Ethylene glycol dissolved= $\text{2075g}$

Boiling point of Ethylene glycol = $\text{197°C}$

Conversion of grams of ethylene glycol to moles

$\frac{{\text{685g C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}}{\text{62}\text{.07g/mol}}\text{= 11}\text{.04 mol}$

By plugging in the value of mass and molar mass of ethylene glycol, the number of moles of ethylene glycol has calculated.

Calculation of molality of the solution

$\frac{\text{11}{\text{.04mol C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{O}}_{\text{2}}}{\text{2}\text{.075kg water}}\text{= 5}\text{.32 m}$

By plugging in the value of moles of ethylene glycol and amount of water, the molality of the solution has calculated.

Calculation of molality of glucose solution

$\text{1062 g solution -170}\text{.1g glucose = 892g water = 0}\text{.892kg water}$

$\frac{\text{0}\text{.9440mol}}{\text{0}\text{.892kg}}\text{= 1}\text{.06m}$

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

Calculation of boiling point of the solution

${\text{ΔT}}_{\text{b}}{\text{ = K}}_{\text{b}}\text{m =(0}\text{.52°C/m)(5}\text{.32m) = 2}\text{.8°C}$

The boiling point of the solution is (100+ $\text{2}\text{.8°C)}$= $\text{102}\text{.8°C}$

By plugging in the value of freezing point and molal freezing point depression constant for ethylene glycol, freezing point of the solution has calculated.

Conclusion

Boiling point of the solution has calculated as $\text{102}\text{.8°C}$