Chapter 13.4, Problem 2CYU

What quantity of ethylene glycol, HOCH₂CH₂OH, must be added to 125 g of water to raise the boiling point by 1.0 °C? Express the answer in grams.

Interpretation Introduction

Interpretation: The quantity of ethylene glycol added in the water to raise the boiling point by $\text{1}{\text{.0}}^{\text{o}}\text{C}$ has to be identified.

Concept introduction:

Boiling point elevation: The expression that depicts the relationship between boiling point of the solution varies with the solute concentration is shown as follows,

  $\text{Elevation}\text{ in boiling point = }\Delta {\text{T}}_{bp}{\text{= K}}_{bp}{\text{. m}}_{solute}$

Here, ${\text{K}}_{bp}$ is molal boiling point elevation constant.

    ${\text{m}}_{solute}$is molality of the solute.

Answer to Problem 2CYU

The quantity of ethylene glycol added in water is$\text{15}\text{g}$

Explanation of Solution

Given:

  $\begin{array}{c}\text{Mass of Water (solvent) =}\text{125 g }\\ \\ \text{= 0}\text{.125 kg }\\ \\ \text{Boiling}\text{point}\text{of}\text{water}{\text{increased by, ΔT}}_{\text{bp}}\text{ = 1}{\text{.0}}^{\text{o}}\text{C}\\ \\ \text{Mass of Ethylene glycol (solute) = ?}\end{array}$

  $\begin{array}{l}\underset{¯}{\text{Boiling point elevation:}}\\ \text{Elevation}{\text{ in boiling point = ΔT}}_{\text{bp}}{\text{= K}}_{\text{bp}}{\text{. m}}_{\text{solute}}\text{,}\end{array}$

From the above equation, mass of the solute can be calculated.

  $\begin{array}{c}\text{1}{\text{.0}}^{\text{o}}\text{C}\text{=}\left(\text{+0}\text{.5121}{}^{\text{o}}\text{C/m}\right)\left({\text{m}}_{\text{solute}}\right)\\ \\ {\text{m}}_{\text{solute}}\text{=}\frac{\text{1}\text{.0}{}^{\text{o}}\text{C}}{\text{+0}\text{.5121}{}^{\text{o}}\text{C/m}}\\ \\ {\text{m}}_{\text{solute}}\text{ = 1}\text{.95}\text{m}\\ \\ \left[{\text{Here, K}}_{\text{bp}}\text{constant value for water}\text{=+0}\text{.5121}{}^{\text{o}}\text{C/m}\right]\end{array}$

Number of moles of glycol can be calculated as given below:

$\begin{array}{c}\text{Moles of solute (mol) = }\left(\text{molality}\right)\left[\text{Amount of solvent}\left(\text{in kg}\right)\right]\\ \\ \text{ = }\left(1.95\text{m}\right)\left(\text{0}\text{.125 kg}\right)\\ \\ \text{ = 0}\text{.24 mol}\text{.}\end{array}$

From the general equation, $No.ofmoles=\frac{mass}{molarmass}$

  $\begin{array}{c}\text{Mass}\text{of}\text{Ethylene}\text{glycol =}\left(\text{No}\text{.of}\text{moles}\right)\left(\text{Molar}\text{mass}\text{of}\text{glycol}\right)\\ \\ \text{=}\left(\text{0}\text{.24 mol}\right)\left(\text{62}\text{.07}\text{g/mol}\right)\\ \\ \text{=}\text{15}\text{g}\end{array}$

Conclusion

The quantity of ethylene glycol required to be added in water was identified.