Question

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What are the mole fraction and the molality of a solution that contains 0.850 g of ammonia, NH₃, dissolved in 125 g of water?

Short answer

The mole fraction and molality of a solution is 7.15 × 10^-3 and 0.399 m.

Step-by-step solution

Mole Fraction and Molality

A mole fraction is a unit of concentration, defined to be equal to the number of moles of a component divided by the total number of moles of a solution.

\{\bf{No}}{\bf{. of moles = }}\frac{{{\bf{Mass of solute}}}}{{{\bf{Molar mass}}}}\)

Molality is defined as the “total moles of a solute contained in a kilogram of a solvent.” Molality is also known as molal concentration.

\{\bf{Molality = }}\frac{{{\bf{Number of Moles}}}}{{{\bf{Mass of Solvent}}\,{\bf{(Kg)}}}}\)

Explanation

By using the above data for the masses of ammonia and the solvent, we can find out the molarity. By using the number of moles, we can find the mole fraction.

\(\begin{aligned}{}{\rm{Mass of N}}{{\rm{H}}_{\rm{3}}}{\rm{ = 0}}{\rm{.850g}}\\{\rm{Molar Mass o}}{{\rm{f}}^{}}{\rm{N}}{{\rm{H}}_{\rm{3}}}{\rm{ = }}\frac{{{\rm{174g}}}}{{{\rm{mole}}}}\\{\rm{Number of Moles of N}}{{\rm{H}}_{\rm{3}}}{\rm{ = }}\frac{{{\rm{0}}{\rm{.850g }}}}{{{\rm{ 17gmol}}{{\rm{e}}^{{\rm{ - 1}}}}}}\\{\rm{Number of Moles of N}}{{\rm{H}}_{\rm{3}}}{\rm{ = 0}}{\rm{.05mole}}\end{aligned}\)

Mass of Solvent, Water = 125 g

Molar Mass of Solvent = 18 g/mole

\(\begin{aligned}{c}{\rm{Number Moles of Solvent = }}\frac{{{\rm{125g }}}}{{{\rm{18 gmol}}{{\rm{e}}^{{\rm{ - 1}}}}}}\\{\rm{ = 6}}{\rm{.94mole}}\\{\rm{Mole Fraction of N}}{{\rm{H}}_{\rm{3}}}{\rm{ = }}\frac{{{\rm{0}}{\rm{.05mole}}}}{{{\rm{0}}{\rm{.05mole + 6}}{\rm{.94mole }}}}\\{\rm{ = 7}}{\rm{.15 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\end{aligned}\)

Mass of Solvent = 0.125 kg

Number of moles of \({\rm{N}}{{\rm{H}}_{\rm{3}}}\)= 0.05 mole

\({\rm{Molality = }}\frac{{{\rm{0}}{\rm{.005}}}}{{{\rm{0}}{\rm{.125}}}}{\rm{ = 0}}{\rm{.399m }}\)