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Chapter 11: Q4CYL (page 618)

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 掳C is 23.7 tor, what is the vapour pressure of the solution?

Short Answer

Expert verified

Thevapour pressure of the solutionis 23.4 tor.

Step by step solution

01

Vapour Pressure

Vapour pressure may be defined as the pressure which will be exerted by the vapour.

Here, Raoult鈥檚 Law may be defined as

\({{\bf{P}}_{\bf{A}}} \propto {{\bf{X}}_{\bf{A}}}\)

\({{\bf{P}}_{{\bf{solution}}}}{\bf{ = }}{{\bf{X}}_{{\bf{solvent}}}}{{\bf{P}}_{{\bf{solvent}}}}\)

02

Explanation

\(\)

Mass of Urea,\({\rm{CO}}{\left( {{\rm{N}}{{\rm{H}}_{\rm{2}}}} \right)_{\rm{2}}}\)= 5.00 g

Molar Mass of Urea,\({\rm{CO}}{\left( {{\rm{N}}{{\rm{H}}_{\rm{2}}}} \right)_{\rm{2}}}\)= 60 g/mole

\({\rm{Number of Moles of Urea = }}\frac{{{\rm{5}}{\rm{.00g}}}}{{{\rm{60gmol}}{{\rm{e}}^{{\rm{ - 1}}}}}}{\rm{ = 0}}{\rm{.083mole}}\)

Mass of Water = 0.100 kg = 100 g

Molar Mass of Water = \({\rm{18 gmol}}{{\rm{e}}^{{\rm{ - 1}}}}\)

\({\rm{Number of Moles of Water = }}\frac{{{\rm{100g}}}}{{{\rm{18 gmol}}{{\rm{e}}^{{\rm{ - 1}}}}}}{\rm{ = 5}}{\rm{.55 mole}}\)

\(\)

\(Mole{\rm{ }}Fraction{\rm{ }}of{\rm{ }}Water = {\rm{ }}\frac{{Mole{\rm{ }}of{\rm{ }}Water{\rm{ }}}}{{Total{\rm{ }}Number{\rm{ }}of{\rm{ }}Moles}}\)

\({\rm{Mole Fraction of Water = }}\frac{{{\rm{5}}{\rm{.55}}}}{{\left( {{\rm{5}}{\rm{.55 + 0}}{\rm{.083}}} \right)}}{\rm{ = 0}}{\rm{.990}}\)

According to Raoult鈥檚 Law,

\(\begin{aligned}{{}{}}{{{\rm{P}}_{{\rm{solution}}}}{\rm{ = }}{{\rm{X}}_{{\rm{solvent}}}}{{\rm{P}}_{{\rm{solvent}}}}}\\{{{\rm{P}}_{{\rm{solution}}}}{\rm{ = 0}}{\rm{.990 \times 23}}{\rm{.7 = 23}}{\rm{.4tor}}}\end{aligned}\)

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