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Chapter 11: Q5CYL (page 619)

What is the boiling point of the antifreeze 17.9m Ethylene Glycol?

Short Answer

Expert verified

The boiling point of the antifreeze 17.9m Ethylene Glycol is 109.2oC.

Step by step solution

01

Boiling Point

At standard atmospheric pressure, the boiling may be defined as the temperature at which the liquid is converted into vapours.

\({\bf{\Delta }}{{\bf{T}}_{\bf{b}}}{\bf{ = }}{{\bf{K}}_{\bf{b}}}{\bf{m}}\)

where,

\({{\bf{K}}_{\bf{b}}}\)= Base Dissociation Constant

m = Molality of the solute

02

Explanation

\(\)

The molality of Ethylene Glycol = 17.9m

Calculate the change in boiling point:

\({\rm{\Delta }}{{\rm{T}}_{\rm{b}}}{\rm{ = }}{{\rm{K}}_{\rm{b}}}{\rm{m = 0}}{\rm{.52 \times 17}}{\rm{.9 = 9}}{\rm{.}}{{\rm{2}}^{\rm{o}}}{\rm{C}}\) \(\)

This adds a boiling point elevation to the boiling point of a pure solvent.

\({\rm{Boiling point temperature = 10}}{{\rm{0}}^{\rm{o}}}{\rm{C + 9}}{\rm{.}}{{\rm{2}}^{\rm{o}}}{\rm{C = 109}}{\rm{.}}{{\rm{2}}^{\rm{o}}}{\rm{C}}\)

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Most popular questions from this chapter

A solution contains 5.00 g of urea, \({\bf{CO}}{\left( {{\bf{N}}{{\bf{H}}_{\bf{2}}}} \right)_{\bf{2}}}\)(a non-volatile solute) and 0.100 kg of water. If the vapour pressure of pure water at 25 °C is 23.7 tor, what is the vapour pressure of the solution?

What is the difference between a 1 M solution and a 1 m solution?

Indicate the most important type of intermolecular attraction responsible for solvation in each of the following solutions:

  1. the solutions in Figure

2. Methanol,\({\bf{CH}}3{\bf{OH}}\), dissolved in ethanol, \({\bf{C}}2{\bf{H}}5{\bf{OH}}\)

3. Methane,\({\bf{C}}{{\bf{H}}_{\bf{4}}}\), dissolved in benzene, \({{\bf{C}}_{\bf{6}}}{{\bf{H}}_{\bf{6}}}\)

4. The polar halocarbon \({\bf{C}}{{\bf{F}}_{\bf{2}}}{\bf{C}}{{\bf{l}}_{\bf{2}}}\)dissolved in the polar halocarbon \({\bf{CF}}2{\bf{ClCFCl2}}\)

5. \({{\bf{O}}_{\bf{2}}}\left( {\bf{l}} \right)\)in \({{\bf{N}}_{\bf{2}}}\left( {\bf{l}} \right){\bf{.}}\) \(\)

Arrange the following solutions in order by their decreasing freezing points: 0.1 m \({\bf{Na}}{}_3{\bf{P}}{{\bf{O}}_4}\), 0.1 m \({{\bf{C}}_2}{{\bf{H}}_5}{\bf{OH}}\), 0.01 m \({\bf{C}}{{\bf{O}}_2}\), 0.15 m NaCl, and 0.2 m \({\bf{CaC}}{{\bf{l}}_2}\)

Assume that each of the ions in calcium chloride, \({\bf{CaC}}{{\bf{l}}_{\bf{2}}}\), has the same effect on the freezing point of water as a nonelectrolyte molecule. Calculate the freezing point of a solution of 0.724 g of \({\bf{CaC}}{{\bf{l}}_{\bf{2}}}\)in 175 g of water.
See all solutions

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