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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

Expert verified

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

Step by step solution

01

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)}}}}\)

02

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

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

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Most popular questions from this chapter

Why does 1 mole of sodium chloride depress the freezing point of 1 kg of water almost twice as much as 1 mole of glycerine?

A 12.0-g sample of a nonelectrolyte is dissolved in 80.0 g of water. The solution freezes at −1.94 °C. Calculate the molar mass of the substance. The Vapour Pressure of the Solution is 23.35 Tor.

Exposing a 100.0 mL sample of water at 0 °C to an atmosphere containing a gaseous solute at 20.26 kPa (152 torr) resulted in the dissolution of\({\bf{1}}.{\bf{45}}{\rm{ }} \times {\rm{ }}{\bf{1}}{{\bf{0}}^{ - {\bf{3}}}}\))g of the solute. Use Henry’s law to determine the solubility of this gaseous solute when its pressure is 101.3 kPa (760 torr).

Calculate the mole fraction of each solute and solvent:

  1. 583 g of\({{\bf{H}}_2}{\bf{S}}{{\bf{O}}_4}\)in 1.50 kg of water—the acid solution used in an automobile battery
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  3. 46.85 g of codeine, \({{\bf{C}}_{{\bf{18}}}}{{\bf{H}}_{{\bf{21}}}}{\bf{N}}{{\bf{O}}_{\bf{3}}}\), in 125.5 g of ethanol, \({{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{5}}}{\bf{OH}}\)
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Calculate the molality of each of the following solutions:

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