Calculate the molality, molarity, and mole fraction of NH₃ in ordinary household ammonia, which is an 8.00 mass % aqueous solution (d = 0.9651 g/mL).

Respuesta :

Mole fraction exists a unit of concentration. In the solution, the relative amount of solute and solvents exists measured by the mole fraction and it exists represented by “X.”

Molarity = 4.54 M

Molality = 5.11 m

mol fraction NH₃ = 4.54 mol / 53.87 mol = 0.084

mol fraction [tex]$H_2O[/tex] = 49.33 mol / 53.87 mol = 0.916

mol fraction contains no units.

How to find the molality, molarity, and mole fraction of NH₃?

Take 1.0 L of this solution

Mass = 965.1 g

Mass of NH₃ = 8/100 [tex]*[/tex] 965.1g = 77.21 g

Molar mass NH₃ = 17 g/mol

mol NH₃ in 77.21 g = 77.21g / 17g/mol

= 4.54 mol in 1.0 L solution

Molarity = 4.54 M

You contain 4.54 mol NH₃ dissolved in 965.1g - 77.21 g

= 887.89 g of water

mol NH₃ dissolved in 1.0 kg water = 4.54 mol [tex]*[/tex] 1000 ml/kg / 887.89 g

= 5.11 mol

Molality = 5.11 m

Mol [tex]$H_2O[/tex] in 887.89 g = 887.89 g/18 g/mol = 49.33 mol [tex]$H_2O[/tex]

Mol NH₃ = 4.54

Total moles = 49.33 + 4.54 = 53.87 mol

mol fraction NH₃ = 4.54 mol / 53.87 mol = 0.084

mol fraction [tex]$H_2O[/tex] = 49.33 mol / 53.87 mol = 0.916

mol fraction contains no units.

To learn more about mol fraction refer to:

https://brainly.com/question/14783710

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