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Calcular las fracciones molares de los componentes de una disolución formada por: 250 g de agua, 50 g de ácido acético y 25 g de cloruro de sodio.

User Tmaximini
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The question is: Calculate the molar fractions of the components of a solution made up of: 250 g of water, 50 g of acetic acid and 25 g of sodium chloride.

Answer: The mole fraction of water, acetic acid and sodium chloride is 0.916, 0.054, and 0.028.

Step-by-step explanation:

As number of moles is the mass of substance divided by its molar mass.

Therefore, moles of given species are calculated as follows.

The number of moles of water (molar mass = 18 g/mol):


Moles = (mass)/(molar mass)\\= (250 g)/(18 g/mol)\\= 13.88 mol

The number of moles of acetic acid (molar mass = 60.052 g/mol):


Moles = (mass)/(molar mass)\\= (50 g)/(60.052 g/mol)\\= 0.832 mol

The number of moles of sodium chloride (molar mass = 58.44 g/mol):


Moles = (mass)/(molar mass)\\= (25 g)/(58.44 g/mol)\\= 0.427 mol

Therefore, mole fraction of each component is calculated as follows.

Mole fraction of water:


Mole fraction = (moles of water)/(Total moles)\\= (13.88)/(13.88 + 0.832 + 0.427)\\= (13.88)/(15.139)\\= 0.916

Mole fraction of acetic acid:


Mole fraction = (moles of water)/(Total moles)\\= (0.832)/(13.88 + 0.832 + 0.427)\\= (0.832)/(15.139)\\= 0.054

Mole fraction of NaCl:


Mole fraction = (moles of water)/(Total moles)\\= (0.427)/(13.88 + 0.832 + 0.427)\\= (0.427)/(15.139)\\= 0.028

Thus, we can conclude that mole fraction of water, acetic acid and sodium chloride is 0.916, 0.054, and 0.028.

User Sajidur Rahman
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